如何将char *分配给char数组？

Question

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]);

Answer 1

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;
    }
}

Answer 2

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 .

Answer 3

#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 (ie a null terminated sequence of characters), then there are more specialist functions for that as well. strcpy is one of those functions.

Answer 4

您的分配是错误的，因为不能将char *分配给char数组，而不是使用此分配，则可以使用strcpy（）。