CODEWITHSUNDEEP
c++char
Augusto Picciani
Augusto Picciani

Reputation: 804

How do I assign a char* to a char array?

Compiler tell me "incompatibles type in assignments of char* to char[32]" this is my code:

char* generalOperations[2]={"Off","On"};

 void test(){
   char value[32];
    switch(swapVariable){
     case 0:
      value=generalOperations[0]; //<==Error HERE!
     break;
    }

 }

[Solved]:

  strcpy(value,generalOperations[0]);

Upvotes: 1

Views: 493

Answers (4)

hmjd
hmjd

Reputation: 121961

You can't assign arrays. You can either change the type of value to a char* or copy the content of generalOptions[0] into value. If you are going to copy the content, then you need to ensure that value has enough space to hold the content of the element in generalOperations.

Modifying a string literal is undefined behaviour, by changing the type to const char* the compiler can detect any attempt to modify one of the entries in generalOperations instead of experiencing odd behaviour at runtime:

const char* generalOperations [2]={"Off","On"};

const char* value;

Note you don't have to specify the number of elements in the array if you are initialising it:

const char* generalOperations [] = {"Off","On"};

Or, if this really is C++ you can make value a std::string instead and just assign to it which will copy the generalOperations element.

As C++ appears to really be the language and C++11 features are permitted instead of using a switch you could create a std::map that associates the int values with the std::string:

#include <iostream>
#include <string>
#include <map>

const std::map<int, std::string> generalOperations{ {17, "Off"},
                                                    {29, "On" } };
int main()
{
    auto e = generalOperations.find(17);
    if (e != generalOperations.end())
    {
        // Do something with e->second.
        std::cout << e->second << "\n";
    }
    return 0;
}

Demo: http://ideone.com/rvFxH.

Upvotes: 3

dakn&#248;k
dakn&#248;k

Reputation: 646

Use std::string instead of char* and std::array<T, N> instead of T[N]. Both are type safe (as opposed to memcpy), both are in modern C++ style and both are directly assignable using the assignment operator.

#include <array>
#include <string>

std::array<std::string, 2> generalOperations{"Off", "On"};

void test() {
    std::string value;
    switch(swapVariable) {
        case 0: value = generalOperations[0]; break;
    }
}

Upvotes: 8

AAA
AAA

Reputation: 448

Your assignment is wrong, since you cannot assign a char * to char array instead of using this assignment you can use strcpy().

Upvotes: -1

john
john

Reputation: 8027

#include <string.h>

...
strcpy(value, generalOptions[0]);

You cannot assign arrays in C/C++. There are functions do to that for you. If your char array represents a C style string (i.e. a null terminated sequence of characters), then there are more specialist functions for that as well. strcpy is one of those functions.

Upvotes: 0

Related Questions