strcpy pass initialized null pointer c [duplicate]

问题

I'm trying out the following code:

int main()
{
   char *yo = "yo";
   char *whaddup = NULL;
   strcpy(whaddup,yo);
}

and I get a segfault. Complete C noob here - other places say I should initialize whaddup as an array. Why can't I pass in a pointer to null?

回答1:

Just any strcpy documentation will tell you that the destination string should be a char array large enough to hold the string and the NUL terminator.

So what you need is something like

char* yo = "yo";
char* whaddup = malloc(strlen(yo)+1);
strcpy(whaddup, yo);

Or alternatively you could use strdup function which does this for you, but it's not standard as it's POSIX only.

回答2:

You have to either declare whaddup as a character array or ideally allocate space for it with malloc.

int main()
{
   char *yo = "yo";
   char *whaddup = malloc(sizeof(char) * 8); // buffer for 8 characters
   strcpy(whaddup,yo);
}

By initializing whaddup to NULL you are not giving it any space in memory, so even copying one character into it will result in a segmentation fault.

回答3:

You can pass in a pointer to null, but you can not copy string from pointer to null.

The function char* strcpy(char* dest, const char* src) is copy string from address src to address dest, your dest is null.

来源：https://stackoverflow.com/questions/38065778/strcpy-pass-initialized-null-pointer-c

标签

strcpy