如何在 C++ 中返回本地数组?

2022-01-12 00:00:00 char c++ return-type 
char *recvmsg(){
    char buffer[1024];
    return buffer;
}

int main(){
    char *reply = recvmsg();
    .....
}

我收到警告:

警告 C4172:返回局部变量或临时地址

warning C4172: returning address of local variable or temporary

推荐答案

你需要动态分配你的char数组:

You need to dynamically allocate your char array:

char *recvmsg(){
   char* buffer = new char[1024];
   return buffer;
}

对于 C++ 和

char *recvmsg(){
   char* buffer = malloc(1024);
   return buffer;
}

对于 C.

如果没有动态分配，您的变量将驻留在函数的堆栈中，因此会在退出时被销毁.这就是你收到警告的原因.在堆上分配它可以防止这种情况发生，但是您必须小心并通过 delete[] 完成后释放内存.

What happens is, without dynamic allocation, your variable will reside on the function's stack and will therefore be destroyed on exit. That's why you get the warning. Allocating it on the heap prevents this, but you will have to be careful and free the memory once done with it via delete[].

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