Statements with procedures

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This page has some examples and questions involving assignment statements with procedures.

Note that all procedures use call by value for input values and call be reference for a value to be changed. The call then acts as a statement that is called for effect (to change the actual variable passed by reference).

Here is the C code.

#include <stdio.h>

void v(int w, int *q) {
	*q = 6 * w;
	}

void t(int b, int u, int *n) {
	*n = b + u;
	}

void k(int g, int *s) {
	*s = 4 / g;
	}

int main() {
	int h, e, z, x, c;
	h = 3;
	e = 7;
	v(e, &z);
	t(h, z, &x);
	k(x, &c);
	printf("%d",c % 10);
	return 0;
	}

Here are the final values of the variables in the program (main).

h == 3 e == 7 z == 42 x == 45 c == 0

The final value of variable c is 0. The output is the remainder left over when c is divided by 10, which is 0 (the answer). Here is the output of the C code.

0

Here is the C code.

#include <stdio.h>

void c(int v, int *p) {
	*p = v + 7;
	}

void s(int z, int f, int *r) {
	*r = z * f;
	}

void t(int b, int *d) {
	*d = 5 / b;
	}

int main() {
	int g, a, u, h, w;
	g = 3;
	a = 4;
	c(g, &u);
	s(u, a, &h);
	t(h, &w);
	printf("%d",w % 10);
	return 0;
	}

Here are the final values of the variables in the program (main).

g == 3 a == 4 u == 10 h == 40 w == 0

The final value of variable w is 0. The output is the remainder left over when w is divided by 10, which is 0 (the answer). Here is the output of the C code.

0

Here is the C code.

#include <stdio.h>

void p(int q, int *b) {
	*b = 6 / q;
	}

void w(int x, int *z) {
	*z = 7 * x;
	}

void h(int c, int d, int *a) {
	*a = c + d;
	}

int main() {
	int r, y, e, m, u;
	r = 5;
	y = 4;
	p(r, &e);
	w(y, &m);
	h(e, m, &u);
	printf("%d",u % 10);
	return 0;
	}

Here are the final values of the variables in the program (main).

r == 5 y == 4 e == 1 m == 28 u == 29

The final value of variable u is 29. The output is the remainder left over when u is divided by 10, which is 9 (the answer). Here is the output of the C code.

9