Output of a c code like with call by reference [closed]

Answer 1

Since there's no recursion here, and all variables have distinct names, we can "cheat" a bit and change the local variables to globals (since this won't cause any name conflicts):

int x=12, y=10; /* original globals */
int tmp;        /* belongs to tswap */
int a;          /* belongs to main */

void tswap(int pa, int pb) {
   tmp=pa;
   pa=pb;
   pb=tmp;
   x=x+pa;
   x=x-pb;
   y++;
   printf("%d %d %d %d\n",pa,pb,x,y);
}

int main() {
    a=4;
    tswap(x,a);
    printf("%d %d %d\n",x,y,a);
    return 0;
}

Since tswap is only called once, we know that its pa parameter is always aliased to x, and its pb parameter is always aliased to a. So, we can just get rid of the parameters, replacing them with the variables that they alias:

int x=12, y=10;
int tmp;
int a;

void tswap() {
   tmp=x;
   x=a;
   a=tmp;
   x=x+x;
   x=x-a;
   y++;
   printf("%d %d %d %d\n",x,a,x,y);
}

int main() {
    a=4;
    tswap();
    printf("%d %d %d\n",x,y,a);
    return 0;
}

At this point, this is just regular C code; we no longer have any parameters, so "pass-by-reference" and "pass-by-value" are equivalent. Tracing through:

int x=12, y=10;
int tmp;
int a;

void tswap() {
   tmp=x;       /* x=12, y=10, tmp=12, a=4 */
   x=a;         /* x=4, y=10, tmp=12, a=4 */
   a=tmp;       /* x=4, y=10, tmp=12, a=12 */
   x=x+x;       /* x=8, y=10, tmp=12, a=12 */
   x=x-a;       /* x=-4, y=10, tmp=12, a=12 */
   y++;         /* x=-4, y=11, tmp=12, a=12 */
   printf("%d %d %d %d\n",x,a,x,y);    /* prints "-4 12 -4 11" */
}

int main() {
    a=4;        /* x=12, y=10, tmp is uninitialized, a=4 */
    tswap();    /* x=-4, y=11, tmp=12, a=4; prints "-4 12 -4 11" */
    printf("%d %d %d\n",x,y,a); /* prints "-4 11 4" */
    return 0;
}