What is the output for the code -
void main()
[
int i=4,x;
x=++i_+++i+_++i;
printf("%d",x);
]
Ashish_16 0 Newbie Poster
rubberman 1,355 Nearly a Posting Virtuoso Featured Poster
What's with the underscores? Is that a C11 thing? Ignoring the sytactical errors, the order of evaluation is compiler-specific. Until I see some valid (in my eye) code, I cannot answer your question. If what you meant was x = ++i + ++i + ++i; then anwser should be either 18 or 15. What does your system show? Mine returns 19 for x. If I change the code to this:
#include <stdio.h>
int main(void)
{
int i=4,x;
int i1 = ++i;
int i2 = ++i;
int i3 = ++i;
x = i1 + i2 + i3;
// x=++i + ++i + ++i; // Returns 19 as x.
printf("i1 == %d, i2 == %d, i3 == %d, x == %d\n",i1, i2, i3, x); // Returns 18 as x, 5 as i1, 6 as i2, and 7 as i3.
return 0;
}
The result of x is 18, as expected.
The lesson here is to NOT have functions (including increment operators) that operate on the same variable within a single expression. This will cause you more problems than you can imagine. I have spent days debugging these problems for colleagues in the past when their code was seriously broken.
If I just set x to ++i, the expected result of i == 5 and x == 5 is output. If I set x to ++i + ++i, the result is i == 6 (expected) but x is 12. This is telling me that the original value of I is used for each increment, not the incremented value of the other operand of the expression, resulting in a value of 12 instead of the expected 11 for x. This is not wrong since these decisions are compiler-dependent. Caveat programmer!
Note that I am using GCC 4.4.7.
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