Thread: using #if debug 1

i have some code that maybe able to be reused if i can add an extra for loop

Code:

#include <stdio.h>
#include <stdlib.h>
#define debug 0
int main()
{
    int i, j, x, y =3;
    if (y == 3)
    {
        #undef debug
        #define debug 1
#if debug 1
    }
    for (j = 0 j< 2; j++)
    {
 #endif // debug
        for (1 = 0; i < 5; i++)
        {
            x += 1;
        }
#if debug 1
    }
#endif // debug
    return 0;
}

i know this is badly written but the principle is if y is 3 add the extra for loop so x will be 10. if y isnt 3 then just use the inner loop ( x will be 5)

is there away to do this that doesn't cause umpteen errors

Define the symbol at the command line:

Code:

$ gcc -DDEBUG -g -o test test.c

Or NDEBUG, since functions as assert() use this symbol to disable debugging... So, instead of #ifdef you can use #ifndef.

debug was the wrong use in that sense. i was trying to turn a for loop on and off so that i can reuse a function rather having the same code repeated with an extra for loop ie

Code:

#include <stdio.h>
#include <stdlib.h>

void func1(void);
void func2(void);
int main()
{
    int y =3;
    if (y == 3)
    {
        func2();
    }
    else
    {
        func1();
    }
    return 0;
}
void func1(void)
{
    int i, x;
    for (i=0;i<=5; i++)
    {
        x += 1;
    }
    printf("%d",x);
}
void func2(void)
{
    int i, j, x;
    for (j=0;j<=2;j++)
    {
        for (i=0;i<=5; i++)
        {
            x += 1;
        }
    }
    printf("%d",x);
}

hope this is clearer
coop

so i could do something like

Code:

#include <stdio.h>
#include <stdlib.h>

int func1(void);
void func2(void);
int main()
{
    int x=0, y =3;
    if (y == 3)
    {
        func2();
    }
    else
    {
       x= func1();
       printf("%d",x);
    }
    return 0;
}
int func1(void)
{
    int i, x=0;
    for (i=0;i<=5; i++)
    {
        x += 1;
    }
    return x;
}
void func2(void)
{
    int j, x=0;
    for (j=0;j<=2;j++)
    {
        x += func1();
    }
    printf("%d",x);
}