Hello guys, I want to alter a string inside a function, but it is not working.
Here is the function:
Code:#include <stdio.h> void test (char *ch){ ch[0] = 0; ch[1] = 1; ch[2] = '\0'; } int main(){ int i; char ch[20]; test(ch); for(i=0;i<20 && ch[i] != '\0';i++){ printf("%c", ch[i]); } return 0; }
> ch[0] = 0;
> ch[1] = 1;
Since 0 is the same as '\0', I'd say it works perfectly and doesn't print anything at all.
Perhaps you expected to see "01" printed, in which case you need
ch[0] = '0';
ch[1] = '1';
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.
Yes, indeed. The character that causes a zero to be printed does not have a value of zero (i.e. a test "0 == '0'" is always false). That is true for all real-world character sets.
