I'm a novice, but I've put plenty of effort into figuring this out myself, and I'm baffled. When I look up the format for a for loop, it says there should be an "expression" in each of the three spots. an if statement counts as a single expression, no? But this:


1 #include <stdio.h>
2
3 int main(void)
4 {
5 int i;
6
7 for (; i < 10; if (1) i++);
8
9 return 0;
10 }

gives me the error:
forLoopTest.c:7: error: expected expression before ‘if’.

and I found ten different examples of people fixing the error "expected expression before" . . . , and they all looked like totally different problems.

I know it doesn't work, and I can't think of any reason I'd need to use an if statement there, but I'm curious why the if statement doesn't work. And knowing what that error means would actually be useful.

No. an "if statement" is (surprise) a statement. Not an expression.

An expression results in a value, a statement 'does' something.

for (; i < 10; if (1) i++);

The part in bold has to be an expression - not an evaluation criteria per se. That is the syntactical limitation placed by the programming language


you could write the if (1) i++ part inside the for loop itself instead. In other words your code would look like this:


#include <stdio.h>

int main(void)
{
int i;

for (; i < 10; ) {
if (1)
i++;
printf("aaa");

}

return 0;

}



Hope this helps!