execution order of loops in C

this is probably a very noob question but I was what the result of this would be:

int someVariable = 1;

while (callFunction(someVariable));

    if (someVariable = 1) {
        printf("a1");
    } else {
        printf("a2");
    }

callFunction (int i) {
    while (i< 100000000) {
        i++;
    }
    return 0;
}

so when you hit the while loop

while (callFunction(someVariable));

does a thread wait at that loop until it finishes and then to

if(someVariable == 1) {
    printf("a1");
} else {
    printf("a2");
}

or does it skip and move to the if condition, print "a2" and then after the loop has finished goes through the if condition again?

UPDATE: This isn't ment to be valid c code just psuedo, maybe I didn't word it right, basically what I'm trying to figure out is what the differ开发者_运维百科ent between a loop like while (callFunction(someVariable)); is vs

while (callFunction(someVariable)){}

i also changed the bolded part in my code i.e ** int someVariable = 1; **, I was doing an endless loop which wasn't my intention.

The code inside a function is executed sequentially, by a single thread. Even if you send an other thread to your function it will execute it sequentually as well.

This is true to 99% of programming languages now days.

UPDATE

basically what I'm trying to figure out is what the different between a loop like while (callFunction(someVariable)); is vs while (callFunction(someVariable)){}

No practical difference. ; delimits an empty statement, { } is a scope without statements. Any compiler can be expected to produce identical code.

Of course, if you want to do something in each iteration of the loop, { } creates a "scope" in which you can create types, typedefs and variables as well as call functions: on reaching the '}' or having an uncaught exception, the local content is cleaned up - with destructors called and any identifiers/symbols use forgotten as the compiler continues....

ORIGINAL ANSWER

This...

callFunction(int i){
     while (i< 100000000){
         i++;
     }
     return 1;
}

...just wastes a lot of CPU time, if the compiler's optimiser doesn't remove the loop on the basis that it does no externally-visible work - i.e. that there are no side-effects of the loop on the state of anything other that "i" and that that's irrelevant because the function returns without using i again. If always returns "1", which means the calling code...

while (callFunction(someVariable)); 

...is equivalent to...

while (1)
    ;

...which simply loops forever.

Consequently, the rest of the program - after this while loop - is never executed.

It's very hard to guess what you were really trying to do.

To get better at programming yourself - understanding the behaviour of your code - you should probably do one or both of:

• insert output statements into your program so you can see how the value of variables is changing as the program executes, and whether it's exiting loops
• use a debugger to do the same

Your code contains an endless loop before any output:

while (callFunction(someVariable));

Did you mean for the ; to be there (an empty loop), or did you mean something else? Not that it matters: callFunction always returns 1, which is converted into true. (If you really want the loop to be empty, at least put the ; on a separate line where it can be seen.)

If you do get beyond the while (because you modify some code somewhere), the if contains an embedded assignment; it's basically the same as:

if ( (someVariable = 1) != 0 )

Which is, of course, always true. (Most C++ compilers should warn about the embedded assignment or the fact that the if always evaluates to true. Or both.)

If your loop completes (it would be sequentially yes, if you fix it), it will print 'a1', since you're doing an assignment in the if, which will always return 1, which evaluates to 'true' for conditionals.