Re: confusion when comparing char * with a string literal

william wrote:
below is a short piece of code I wrote to testify my understanding of
char *, and array.

#include <stdio.h>

Include string.h to get the prototype for strtok.

int main()
{
char *str=NULL;
char x[]="today is good!";
printf("%s", str);

Huh? You're passing a null pointer to printf and telling it to follow
it? I think you meant to initilaise it to point to x.

str = x;

str=strtok(x," ");

strtok need a char * as it's first argument. x is a constant pointer.

if (str=="today") //<==here is line that confuses me

You cannot directly compare strings in C. You'll need to compare them
element by element. The Standard C library function for this is strcmp
or strncmp.

printf("they equals!\n");
return 0;
}

I printed "str" first, and the console displayed "today". However,
when I try to comapare 'str' with "today", the condition failed!

Exactly speaking, I know that 'str' is a 4 byte pointer of char type,

No. The Standard says nothing about the sizes of pointer types. Nor do
you need to assume a particular size for what you're doing. And it's
not a pointer *of* type char. It's a pointer *to* type char.

So, my question is how do we compare arrays and char *(I found that it
is so commonly used as string)?

strcmp/strncmp/memcmp

one more question: is it true that we have to initialize a char * or
points it to somewhere in memory, or to malloc memory to it before we
can use it?

Yes.

Because I got segfault several times arising from this problem too.

Indeed you'll if you work on pointers that don't point to any valid
object.

.

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