I would like to know how it processed the output.....Code:#include <stdio.h> #include <stdlib.h> int main() { int *a, *s, i; s = a = (int *) malloc(4 * sizeof(int)); for(i = 0; i < 4; i++) { *(a + i) = i * 10; /* a[i] = i * 10 */ } printf("*s++ = %d\n", *s++); /* output: 0 */ printf("(*s)++ = %d\n", (*s)++); /* output: 10 */ printf("*s = %d\n", *s); /* output: 11 */ printf("*++s = %d\n", *++s); /* output: 20 */ printf("++*s = %d\n", ++*s); /* output: 21 */ getchar(); return 0; }
thanks.
It's meant to show you the difference between incrementing a pointer (s++) and incrementing what a pointer points to (*s)++
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
yeah, but how is it worked out ?Originally Posted by Salem
for example,
printf("*s++ = %d\n", *s++);
(*s)++ it increments the pointer, but i can't really see how the output is 0
Last edited by dredre; 05-08-2004 at 03:31 PM.
That's because you've forgotten basic multiplication. Anything multiplied by zero is zero:
Quzah.Code:*(a + i) = i * 10; /* a[i] = i * 10 */
Hope is the first step on the road to disappointment.
Well why don't you print out the whole array (there's only 4 numbers) between each of the print statements you have, so you can see how it changes as well.
If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
If at first you don't succeed, try writing your phone number on the exam paper.
