What's wrong with my strcpy? [closed]

问题

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Closed 6 years ago.

I tried to make strcpy myself. It should work, I even copied and pasted the (almost exact code) from someones post here about strcpy. Both give me a "Segmentation Fault".

char* strcpy(char * destination, const char * source)
{
    while( (*destination++ = *source++) != '\0' ) 
        ;
    return destination;
}

What's wrong with this code?

char* a = "hello";
cout << strcpy(a, "Haha") << endl;

回答1:

You are trying to write to data segment since "hello" is stored there.

Therefore, when you call strcpy you get segmentation fault.

Try:

char a[] = "hello";
cout << strcpy(a, "Haha") << endl;

instead.

EDIT: Inside your strcpy function, after the copy, destination will point to end of the string, you need to return beginning of the string instead.

回答2:

You're trying to overwrite a string literal. That causes undefined behaviour. Declare a as an array instead:

char a[] = "hello";

Your strcpy implementation has a bug, too (assuming the normal semantics). It should return a pointer to the beginning of the destination buffer, not the end.

回答3:

a is a pointer to a string literal:

char* a = "hello";

Trying to modify a string literal is undefined behavior. As Carl suggested, if you initialize an array with the string literal instead it will work:

char a[] = "hello" ;

回答4:

Besides everything mentioned above and below about string literals, you're also returning a pointer to the END of your string, so even if you avoid the segmentation fault, you'll print "".

来源：https://stackoverflow.com/questions/16677015/whats-wrong-with-my-strcpy