Member Avatar for warpstar

My Question is regarding the code written below. If you look at the function called "function"
when 'a' is passed to the function and the function "function" is evaluated. I was wondering what the return value would be for the first pass when (a =2), my questin specifcaly is that do we increment b and c since its post? otherwise if they had been pre increment would the value of the expression 'a1+b+c' be any different ? thanks.

int function( int a1 )
{
      int b = 0;
      int c = 0;
      b++;
      c++;
return ( a1 + b + c );
}

int main() /*main function*/
{
  int a = 2, i;

  for( i=0; i<3; i++ )
    printf( "%d ", function(a) );
  return 0;
}
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  • Latest Post Latest Post by Ancient Dragon

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Short answer to : with the code shown, b and c are local and reinitialized to 0 on every call. The two ++ statements run before the return, so the function returns a1 + 1 + 1. If a is 2 each call prints 4 (so the output will be 4 4 4). 's point about pre- vs post-increment is correct: when the increment is its own statement there is no difference in the final numeric result.

Useful clarifications and gotchas:

  • If the increments and the sum happen in the same expression, pre- and post-increment change what value contributes to the expression. For example, with persistent state (static variables) you will see different sequences depending on b++/c++ versus ++b/++c:
int f_post(int a) {
    static int b = 0, c = 0;
    return a + b++ + c++;   // first call uses old b,c, then increments
}

int f_pre(int a) {
    static int b = 0, c = 0;
    return a + ++b + ++c;   // first call increments then uses the new values
}

  • For loops: the usual for (i = 0; i < n; i++) and for (i = 0; i < n; ++i) behave identically for built-in integers. However, for user-defined iterators (containers, custom iterator types) prefer pre-increment (++it) because post-increment usually constructs a temporary copy and can be less efficient.

  • Never rely on modifying the same scalar object more than once in a single unsequenced expression — that causes undefined behavior. For authoritative details on increment semantics and expression evaluation rules see the C++ reference pages for operator++ and evaluation order: operator++ and evaluation order.

This addresses the immediate output question and gives practical variants to try (persistent state, increments inside the expression, iterators) so the differences become obvious.

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Member Avatar for Ancient Dragon

>>otherwise if they had been pre increment would the value of the expression 'a1+b+c' be any different
No they are the same. pre and post increment only has any real difference if used in an expression of some sort. For example

int a = 1;

foo(++a);
foo(a++);

In the code above the value of a inside function foo() will be different because in the first instance the value of a is first incrmeneted then that value is passed to foo(). In the second instance the current value of a is passed to foo() then incremented after foo() returns to its caller.

Member Avatar for warpstar

This goes the same for a "For Loop"??, the increment expression, if its pre/post wouldnt it be the same since its being evaluated after the statements of the loop have been executed?

>>otherwise if they had been pre increment would the value of the expression 'a1+b+c' be any different
No they are the same. pre and post increment only has any real difference if used in an expression of some sort. For example

int a = 1;

foo(++a);
foo(a++);

In the code above the value of a inside function foo() will be different because in the first instance the value of a is first incrmeneted then that value is passed to foo(). In the second instance the current value of a is passed to foo() then incremented after foo() returns to its caller.

Member Avatar for Ancient Dragon

>>This goes the same for a "For Loop"?
Yes, both these have the same identical result.

int i;
for(i = 0; i < 10; i++)
and 
for(i = 0; i < 10; ++i)

These two, however, are different

int i;
for(i = 0; i++ < 10;)
and 
for(i = 0; ++i < 10;)
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