incompatible types in assignment of const char to char

Next: Type Conversions , Previous: Functions , Up: Top   [ Contents ][ Index ]

23 Compatible Types

Declaring a function or variable twice is valid in C only if the two declarations specify compatible types. In addition, some operations on pointers require operands to have compatible target types.

In C, two different primitive types are never compatible. Likewise for the defined types struct , union and enum : two separately defined types are incompatible unless they are defined exactly the same way.

However, there are a few cases where different types can be compatible:

• Every enumeration type is compatible with some integer type. In GNU C, the choice of integer type depends on the largest enumeration value.
• Array types are compatible if the element types are compatible and the sizes (when specified) match.
• Pointer types are compatible if the pointer target types are compatible.
• Function types that specify argument types are compatible if the return types are compatible and the argument types are compatible, argument by argument. In addition, they must all agree in whether they use ... to allow additional arguments.
• Function types that don’t specify argument types are compatible if the return types are.
• Function types that specify the argument types are compatible with function types that omit them, if the return types are compatible and the specified argument types are unaltered by the argument promotions (see Argument Promotions ).

In order for types to be compatible, they must agree in their type qualifiers. Thus, const int and int are incompatible. It follows that const int * and int * are incompatible too (they are pointers to types that are not compatible).

If two types are compatible ignoring the qualifiers, we call them nearly compatible . (If they are array types, we ignore qualifiers on the element types. 7 ) Comparison of pointers is valid if the pointers’ target types are nearly compatible. Likewise, the two branches of a conditional expression may be pointers to nearly compatible target types.

If two types are compatible ignoring the qualifiers, and the first type has all the qualifiers of the second type, we say the first is upward compatible with the second. Assignment of pointers requires the assigned pointer’s target type to be upward compatible with the right operand (the new value)’s target type.

This is a GNU C extension.

• Windows Programming
• UNIX/Linux Programming
• General C++ Programming
• error: assigning to 'char' from incompat

  error: assigning to 'char' from incompatible type 'const char [2]'

• C and C++ FAQ
• Mark Forums Read
• View Forum Leaders
• What's New?
• Get Started with C or C++
• C++ Tutorial
• Get the C++ Book
• All Tutorials
• Advanced Search

• General Programming Boards
• C++ Programming

incompatible types in assignment of 'char*'

• Getting started with C or C++ | C Tutorial | C++ Tutorial | C and C++ FAQ | Get a compiler | Fixes for common problems

Thread: incompatible types in assignment of 'char*'

Thread tools.

• Show Printable Version
• Email this Page…
• Subscribe to this Thread…
• View Profile
• View Forum Posts

I am trying to compile program given in link C++/Classes and Inheritance - Wikiversity but it gives error Code: #include<iostream> class Dog { private: char name[25]; int gender; int age; int size; bool healthy; public: char* getName() { return name; } int getGender() { return gender; } int getAge() { return age; } int getSize() { return size; } bool isHealthy() { return healthy; } void setHealthy(bool dhealthy) { healthy = dhealthy; } void setName(char* dname) { name = dname; } }; int main() { Dog lucy; std::cout << "lucy's name is " << lucy.getName() << std::endl; std::cout << "lucy's gender is " << lucy.getGender() << std::endl; std::cout << "lucy's age is " << lucy.getAge() << std::endl; std::cout << "lucy's size is " << lucy.getSize() << std::endl; if(lucy.isHealthy()) std::cout << "lucy is healthy" << std::endl; else std::cout << "lucy isn't healthy :(" << std::endl; std::cout << "Now I'm changing lucy abit..." << std::endl; lucy.setHealthy(!(lucy.isHealthy())); lucy.setName("lUCY"); std::cout << "lucy's name is " << lucy.getName() << std::endl; std::cout << "lucy's gender is " << lucy.getGender() << std::endl; std::cout << "lucy's age is " << lucy.getAge() << std::endl; std::cout << "lucy's size is " << lucy.getSize() << std::endl; if(lucy.isHealthy()) std::cout << "lucy is healthy" << std::endl; else std::cout << "lucy isn't healthy :(" << std::endl; return 0; } In member function 'void Dog::setName(char*)': error: incompatible types in assignment of 'char*' to 'char [25]' void setName(char* dname) { name = dname; } ^~~~~ In function 'int main()': warning: ISO C++ forbids converting a string constant to 'char*' [-Wwrite-strings] lucy.setName("lUCY"); ^

Last edited by abhi143; 11-20-2019 at 07:18 AM .

You should really consider using std::string instead of the error prone C-strings. In your setName() function you should first verify that the passed C-string is small enough to fit into the name array, then use strcpy() to copy the passed string into the name array. Also do your realize that you're trying to print your uninitialized class variables on lines 29 thru 36? All of your getXXXX() member functions should be const qualified. Code: ... int getGender() { return gender; } const ... Lastly, for now, you really should find a different source. The one you've found seems to have quite a few deficiencies and is really doing you a disservice. Edit: I wonder why gender is an int instead of either a string or a single char.

Lastly, for now, you really should find a different source. The one you've found seems to have quite a few deficiencies and is really doing you a disservice. Not really, Wikiversity is a really good source to learn from. @abhi143 rewrote the code in a different way, probably for practice, and that's what has produced him his errors. You can see the website for yourself, it's pretty good for someone starting out on classes and inheritance and abstraction. Also do your realize that you're trying to print your uninitialized class variables on lines 29 thru 36? Wikiversity: However, if we run this program, we might have a little problem. We have never initialized lucy's properties... Well, they try and teach about initialization through this. You probably didn't look through the website but that's not your fault. It's a really reliable place to learn from. Code: In member function 'void Dog::setName(char*)': Code: error: incompatible types in assignment of 'char*' to 'char [25]' void setName(char* dname) { name = dname; } ^~~~~ In function 'int main()': warning: ISO C++ forbids converting a string constant to 'char*' [-Wwrite-strings] lucy.setName("lUCY"); ^ You can't set one pointer to another pointer... Use std::string as mentioned (and written) on the website instead of experimenting with C-style strings if you're wanting to learn C++. Or, if you're trying to convert the C++ code to C code, you should do what @jimblumberg suggested.

Originally Posted by jimblumberg You should really consider using std::string instead of the error prone C-strings.. Code: #include<iostream>using namespace std; class Dog { private: char name[25]; int gender; int age; int weight; public: char* getName() { return name; } int getGender() { return gender; } int getAge() { return age; } int getWeight() { return weight; } }; int main() { Dog lucy; cout << "lucy's name is " << lucy.getName() << endl; cout << "lucy's gender is " << lucy.getGender() << endl; cout << "lucy's age is " << lucy.getAge() << endl; cout << "lucy's Weight is " << lucy.getWeight() << endl; return 0; } lucy's name is a lucy's gender is 4201152 lucy's age is 6422368 lucy's Weight is 4201243 How to print correct details of lucy's ?

> How to print details of lucy's ? Exactly the way you do. Again consider using std::string as that's what you should be doing if you want to learn C++. Read a little further in the Wikiversity Guide you're referring to. It does teach you about Constructors and Initialization. Right now, all the details you print are just the stuff existing in uninitialized memory, exactly what @jimblumberg mentioned. You need to complete a topic completely in order to get the simplest program of that topic to work. Learning in bits and pieces won't do you good. Good luck

Originally Posted by Zeus_ > How to print details of lucy's ? Exactly the way you do. Again consider using std::string as that's what you should be doing if you want to learn C++. Read a little further in the Wikiversity Guide you're referring to. It does teach you about Constructors and Initialization. Right now, all the details you print are just the stuff existing in uninitialized memory, exactly what @jimblumberg mentioned. You need to complete a topic completely in order to get the simplest program of that topic to work. Learning in bits and pieces won't do you good. Good luck There is no much information given in link. I created program after spending few time Code: #include<iostream> using namespace std; class Dog { private: int age; char gender; float weight; public: int getAge() { cout << "\nEnter lucy's age: "; cin >> age; return age; } char getGender() { cout << "\nEnter lucy's Gender: "; cin >> gender; return gender; } float getWeight() { cout << "\nEnter lucy's Weight: "; cin >> weight; return weight; } }; int main() { Dog lucy; cout << "lucy's age is " << lucy.getAge() << endl; cout << "lucy's gender is " << lucy.getGender() << endl; cout << "lucy's Weight is " << lucy.getWeight() << endl; return 0; } Enter lucy's age: 5 lucy's age is 5 Enter lucy's Gender: m lucy's gender is m Enter lucy's Weight: 20 lucy's Weight is 20

> There is no much information given in link. There's enough information on there than you've attempted to read. I've looked through myself and everything about initialising properties is present. Look carefully! Read the part under constructors. It may be slightly advanced to you, I feel, and so consider finding a book/guide/online tutorial that explains you classes from the basics. Default Constructors and how writing your own constructor disables the one provided by the compiler, etc etc etc. Additionally you'd also have to learn more on Polymorphism (... Overloading) to get a grasp of the concepts used in most beginner-ish tutorials. Code: class Dog [ edit ] Structs are containers whose properties are always public. In order to gain control over what is public and what isn't, we will have to use a class. Let us see how the Dog class would look like in a class instead of a struct: #include <string> class Dog { private : std :: string name; int gender; int age; int size; bool healthy; }; Now our dogs precious properties aren't public for everyone to view and change. However, this raises a little problem. We can't change them either. Only other dogs can see and change them. In order to be able to access a Dog's private properties, we would have to make a function to do so. This brings us to our next chapter. Dogs with Methods [ edit ] Classes can have functions in them. These functions are called methods. Methods can access all, and even private, properties and methods (yes, methods can be private) of its class. Let us make some methods to get our dog's information... #include <string> class Dog { private : std :: string name; char gender; int age; int size; bool healthy; public : std :: string getName() const { return name; } int getGender() const { return gender; } int getAge() const { return age; } int getSize() const { return size; } bool isHealthy() const { return healthy; } void setHealthy( bool dhealthy) { healthy = dhealthy; } void setName( const std :: string & dname) { name = dname; } }; #include <iostream> int main () { Dog lucy; std :: cout << "lucy's name is " << lucy.getName() << std :: endl; std :: cout << "lucy's gender is " << lucy.getGender() << std :: endl; std :: cout << "lucy's age is " << lucy.getAge() << std :: endl; std :: cout << "lucy's size is " << lucy.getSize() << std :: endl; if (lucy.isHealthy()) std :: cout << "lucy is healthy" << std :: endl; else std :: cout << "lucy isn't healthy :(" << std :: endl; std :: cout << "Now I'm changing lucy abit..." << std :: endl; lucy.setHealthy( ! (lucy.isHealthy())); lucy.setName( "lUCY" ); std :: cout << "lucy's name is " << lucy.getName() << std :: endl; std :: cout << "lucy's gender is " << lucy.getGender() << std :: endl; std :: cout << "lucy's age is " << lucy.getAge() << std :: endl; std :: cout << "lucy's size is " << lucy.getSize() << std :: endl; if (lucy.isHealthy()) std :: cout << "lucy is healthy" << std :: endl; else std :: cout << "lucy isn't healthy :(" << std :: endl; return 0 ; } However, if we run this program, we might have a little problem. We have never initialized lucy's properties... Constructors [ edit ] In order to initialize lucy's properties, we could write a method that would do it for us, and call it every time we make a dog. However, C++ already provides us such a method. A constructor is a method which is called every time a new object is created. To use this nice little thing provided to us by C++, we will have to write a method which returns no value, and has the same name as the class. Here's Dog written using a constructor: #include <string> class Dog { private : std :: string name; char gender; int age; int size; bool healthy; public : std :: string getName() const { return name; } char getGender() const { return gender; } int getAge() const { return age; } int getSize() const { return size; } bool isHealthy() { return healthy; } void setHealthy( bool dhealthy) { healthy = dhealthy; } void setName( const std :: string & dname) { name = dname; } Dog() : name( "Lucy" ), gender( 'f' ), age( 3 ), size( 4 ), healthy( true ) {} }; From now on, every dog we instantiate will have the name "Lucy", be of gender 1, aged 3, sized 4 and healthy. Well, what if you want some diversity? No problem. Constructors with parameters! Dog( const std :: string & dname, int dgender, int dage, int dsize, bool dhealthy) : name(dname), gender(dgender), age(dage), size(dsize), healthy(dhealthy) {} If you have both constructors in your Dog class, you can now either create a dog as always, which will have the default values (named "Lucy" etc etc), or you could use RAII to make your customized dog as follows: #include <iostream> int main () { Dog scruffy( "Scruffy" , 'm' , 8 , 6 , false ); std :: cout << "scruffy's name is " << scruffy.getName() << std :: endl; std :: cout << "scruffy's gender is " << scruffy.getGender() << std :: endl; std :: cout << "scruffy's age is " << scruffy.getAge() << std :: endl; std :: cout << "scruffy's size is " << scruffy.getSize() << std :: endl; if (scruffy.isHealthy()) std :: cout << "scruffy is healthy" << std :: endl; else std :: cout << "scruffy isn't healthy :(" << std :: endl; return 0 ; }

Last edited by Zeus_; 11-21-2019 at 02:52 AM .

Originally Posted by Zeus_ > There's enough information on there than you've attempted to read. I agree with you I have to spent some time on reading that's what I am doing Do you have any advice on my previous code. I wrote it myself so may be somewhere I am wrong

• Visit Homepage

You should not have the member functions do interactive I/O like that.

Originally Posted by Bjarne Stroustrup (2000-10-14) I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool. Look up a C++ Reference and learn How To Ask Questions The Smart Way

Originally Posted by laserlight You should not have the member functions do interactive I/O like that. if possible Can you give any hint How it should be ?

In general, your Class Methods shouldn't be doing interactive I/O. By interactive, @laserlight means you shouldn't be having things like "Enter Lucy's age: " in your method that does I/O. It takes a lot to take understand the "why" because you are indeed learning one of the hardest languages out there for beginners. You'll gradually understand with time the "why" when you start reading about data abstraction etc. For now, you needn't worry about it and should just care about getting things to work and be able to understand them. Make your I/O methods un-interactive. Do the interaction handling separately. Also, you'd probably want the Dog class for different dogs too and not just Lucy. The tutorial explains very well on how to do just that. Also, if seeing things like const written everywhere makes it confusing, you can probably ignore that for now. It's just present so that your methods are valid for const objects too. Your code should be more like: Code: void SetAge () { cin >> Age; // You actually need to check here if the user entered a number and not characters but don't worry about that for now } int GetAge () const { return Age; } And your interaction should be more like: Code: cout << "Enter " << dog.GetName () << " 's age: "; dog.SetAge (); // Assuming "dog" is an instance/object of your Dog class and GetName has been properly defined the way it should be and assuming the name of the dog has been entered already before filling in the age

Last edited by Zeus_; 11-21-2019 at 11:29 AM .

Originally Posted by Zeus_ In general, your Class Methods shouldn't be doing interactive I/O. By interactive, @laserlight means you shouldn't be having things like "Enter Lucy's age: " in your method that does I/O. It takes a lot to take understand the "why" because you are indeed learning one of the hardest languages out there for beginners. You'll gradually understand with time the "why" when you start reading about data abstraction etc. For now, you needn't worry about it and should just care about getting things to work and be able to understand them. yes Practice will make me perfect. I don't get age in method How does it looks Code: #include<iostream>using namespace std; class Dog{ private: int age; public: void setAge (int A) { // Set Age age = A; } int getAge() { // Get Age return age; } }; int main(){ Dog lucy; lucy.setAge(5); cout << "lucy's age is " << lucy.getAge() <<endl; return 0; }

• Private Messages
• Subscriptions
• Who's Online
• Search Forums
• Forums Home
• C Programming
• C# Programming
• Game Programming
• Networking/Device Communication
• Programming Book and Product Reviews
• Windows Programming
• Linux Programming
• General AI Programming
• Article Discussions
• General Discussions
• A Brief History of Cprogramming.com
• Contests Board
• Projects and Job Recruitment

• How to create a shared library on Linux with GCC - December 30, 2011
• Enum classes and nullptr in C++11 - November 27, 2011
• Learn about The Hash Table - November 20, 2011
• Rvalue References and Move Semantics in C++11 - November 13, 2011
• C and C++ for Java Programmers - November 5, 2011
• A Gentle Introduction to C++ IO Streams - October 10, 2011

Similar Threads

[error] incompatible types in assignment of 'const char [6]' to 'char [80]', incompatible types in assignment, incompatible types in assignment., tags for this thread.

View Tag Cloud

• C and C++ Programming at Cprogramming.com
• Web Hosting
• Privacy Statement

incompatible types in assignment of ‘const char [6]’ to ‘CHAR [32]’ {aka ‘char [32]’}

Incompatible types in assignment of 'const char [17]' to 'char [32]', incompatible types in assignment of 'crgb' to 'crgb [42]'.

STM32F0xx产品技术培训.rar

Bsp_uart.rar_stm32f103 uart_bsp_uart_stm32 uart fifo.

Android调试出现The selected device is incompatible问题解决

/home/vrv/src/EDSMClient-XC_svn/MainUI3/switch.cpp:78: 错误: incompatible types in assignment of ‘char*’ to ‘char [10000]’ z_Depart = new char[z_nlen]; ^

[error] incompatible types in assignment of 'student*' to 'student [5]', [error] incompatible types in assignment of 'int*' to 'int [20]', incompatible types in assignment:string,datastore, [error] incompatible types in assignment of 'student*' to 'student [5]'请修改这个错误, error: incompatible types: char cannot be converted to string, error: incompatible types: char[] cannot be converted to char, /home/vrv/src/edsmclient-xc_svn/mainui3/switch.cpp:71: 错误: incompatible types in assignment of ‘qstring’ to ‘char [64]’ m_tempuseraccount=ui->lineedit->text(); ^, 172 18 d:\c\未命名1.cpp [error] incompatible types in assignment of 'student*' to 'student [5]', error: incompatible types in assignment of 'uint8_t* {aka unsigned char*}' to 'uint8_t [4194304] {aka unsigned char [4194304]}' pmonitor->m_roiframe.frame.data=(uint8_t*)roi_img;, error: #167: argument of type "char" is incompatible with parameter of type "const char *", argument of type "const uint8_t *" is incompatible with parameter of type "const char *", argument of type "uint16_t" is incompatible with parameter of type "const char *restrict", assignment to ‘char **’ from incompatible pointer type ‘char (*)是什么意思.

Python零基础30天速通（小白定制版）（完结）

zigbee-cluster-library-specification

实现实时数据湖架构：Kafka与Hive集成

解答下列问题：S—＞S；T｜T；T—＞a 构造任意项目集规范族，构造LR（0）分析表，并分析a;a

Jsbsim reference manual, "互动学习：行动中的多样性与论文攻读经历", 实现实时监控告警系统：kafka与grafana整合, mac上和window原生一样的历史剪切板工具有什么, c++校园超市商品信息管理系统课程设计说明书(含源代码) (2).pdf.

Navigation Menu

Search code, repositories, users, issues, pull requests..., provide feedback.

We read every piece of feedback, and take your input very seriously.

Saved searches

Use saved searches to filter your results more quickly.

To see all available qualifiers, see our documentation .

• Notifications

Have a question about this project? Sign up for a free GitHub account to open an issue and contact its maintainers and the community.

By clicking “Sign up for GitHub”, you agree to our terms of service and privacy statement . We’ll occasionally send you account related emails.

Already on GitHub? Sign in to your account

error: incompatible types in assignment #196

sunnybear commented Oct 25, 2020

anthwlock commented Oct 26, 2020

Sorry, something went wrong.

ambiamber commented Sep 19, 2022

ponchio commented Sep 19, 2022

No branches or pull requests

Getting this error: incompatible types in assignment of 'int' to 'char [16]'

I'm struggling to add a certain string (or char[]) to an array of struct because of the error on the top. I would appreciate it so much if I could get some help with this. The problem is on the line "channels[i].create = Serial.read();".

Serial.read returns an int, but you're trying to assign that int to an array.

Your code lacks a loop function

Ooooh I see. So could I add a loop with an index that goes from 0-25 that has 26 different values?

No, because that array has only 16 elements

Interesting...

My only issue is trying to make it work like:

Do you have any tips?

I don't know what it is you're trying to do. The diagram you posted doesn't seem to relate to the code you posted.

This topic was automatically closed 180 days after the last reply. New replies are no longer allowed.

Related Topics