Function returning wrong value

Question

I don't understand the output from this code:

#include <stdio.h>
long func(long pass)
{
    long ret;
    long i = pass;

    if (i == 6)
    {
        printf("i=%ld\n",i);
        return i;
    }
    printf("ended\n");
}

void main()
{
    int j;
    long it = 0;
    for(j = 0; j < 12; j++)
    {
        printf("%ld\n",func(it));
        it++;
    }
}

The output shows "ended" and "6" every time except when it reaches i=6, that time it prints i=6 and 6.

WHY? It shouldn't be going inside i == 6 every time right?

func doesn't return anything unless pass == 6. That's undefined behavior. — AndyG
– AndyG, Commented Nov 19, 2014 at 2:22
most modern compilers would see there is not every path ends with a return(value) statement and should have raised an error/warning message. Suggest enabling all warnings/errors (for gcc use -Wall) and fixing the problem before trying to run. — user3629249
– user3629249, Commented Nov 19, 2014 at 2:34
Speaking to the title of the question, "Function returning wrong value": When i!=6, what would you consider to be the right value? — DavidO
– DavidO, Commented Nov 19, 2014 at 4:19
gcc gives the following warnings: warning: control reaches end of non-void and warning: return type of 'main' is not 'int'. Read that first — phuclv
– phuclv, Commented Nov 19, 2014 at 8:00

Chris Dodd · Accepted Answer · 2014-11-19 02:24:17Z

3

When i is not equal to 6, you run off the end of the function without a return statement, which gives you undefined behavior -- anything might happen.

In this case it probably just returns whatever happens to be in the hardware return register at the end of the function, which might just be the constant 6, as the compiler might have put it there for the comparison. A different compiler (or even the same compiler run on a different day) might give you a different result.

answered Nov 19, 2014 at 2:24

Chris Dodd

127k14 gold badges150 silver badges243 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

M.M Over a year ago

In C there is no UB if the calling code does not access the returned value (unlike C++)

AndyG Over a year ago

@MattMcNabb: But OP's code is accessing the returned value.

M.M Over a year ago

@AndyG I was responding to first paragraph of Chris's answer

AndyG Over a year ago

@MattMcNabb: I see. Apologies for the confusion.

M.M · Accepted Answer · 2014-11-19 02:24:53Z

2

When i == 6 is not true , then execution reaches the end of func without encountering a return statement. This causes the function to return an indeterminate value.

Then the line printf("%ld\n",func(it)); accesses the indeterminate value, which causes undefined behaviour.

answered Nov 19, 2014 at 2:24

M.M

143k26 gold badges224 silver badges400 bronze badges

Comments

Harminder Singh · Accepted Answer · 2014-11-19 03:53:01Z

0

the control reaches the end of long func(long) because return statement is inside the if condition which fails if 'i' is not equal to 6.

answered Nov 19, 2014 at 3:53

Harminder Singh

11 bronze badge

Comments

PU. · Accepted Answer · 2014-11-19 05:25:16Z

0

When i is not equal to 6 it throws the warning as ::  warning C4715: 'func' : not all control paths return a value
and returns an indeterminate values and printf("%ld\n",func(it)); prints that value like,
ended
1
ended
1
ended
1
ended
1
ended
1
ended
1
i=6
6
ended
6
ended
6
ended
6
ended
6
ended
6

answered Nov 19, 2014 at 5:25

PU.

1489 bronze badges

Collectives™ on Stack Overflow

Function returning wrong value

4 Answers 4

4 Comments

Comments

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

4 Answers 4

4 Comments

Comments

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related