在动态数组上使用strcpy

Question

I have got a basic question about the following code:

int main() {
    char *sNext1="DABCD";
    char Tmp[]="";
    strcpy(Tmp, sNext1);
    return 0;
}

There is a error happened in "strcpy" when I run it. It seems that it doesn't allow me to copy a string into a dynamic array. Does anyone know what causes this error? And how can I fix it?

Answer 1

The problem here is with

 char Tmp[]="";

here, Tmp is not an dynamic array, it's an array with exactly one element (initialized via the initializer). When you use this as destination for strcpy() , you're overrunning the buffer. Basically, your code tries to access out of bound memory, which invokes undefined behavior .

Related, quoting C11 , chapter §7.24.2.3, strcpy() , emphasis mine

The strcpy function copies the string pointed to by s2 (including the terminating null character) into the array pointed to by s1 . [...]

and then, for "String function conventions", §7.24.1,

[....] If an array is accessed beyond the end of an object, the behavior is undefined.

Solution : You need to ensure that Tmp has enough memory to hold the content to be copied into it, including the null terminator.

That said, for a hosted environment, int main() should at least be int main(void) to be conforming to the standard.

Answer 2

Allocate memory to Tmp before copying the string to Tmp .

Tmp is an character array of size 1.

Allocate memory dynamically as shown below. (If you are trying to do dynamic memory allocatio)

char *Tmp = malloc(strlen(sNext1) + 1);

Then try doing strcpy()