C Newbie/ not understanding what might the problem be - c

Question

since im learning C language i decided to make a simple program that adds, substracts and calculates the product of two variables . Depending on users' input whether its 1,2 or 3 to choose addition/substraction/folding.
#include <stdio.h>
int main (void) {
int a, b, c, d, e, f, g;
a=19; b=11; c=a+b; d=a-b; e=a*b; f=-1;
while (f<0 && f>3) {
printf("-press 1 to calculate the sum of a and b\n");
printf("-press 2 to calculate the difference between a and b\n");
printf("-press 3 to calculate the product of a and b\n");
scanf("%d\n",&g);
f=g;
return;
}
if (f == 1) {
printf("A+B= %ls\n", &c);
} else if (f == 2) {
printf("A-B= %ls\n", &d);
} else if (f == 3) {
printf(" A*B= %ls\n", &e);
}
return 0;
}
When i run the program its reads "g" and then it stops.
any suggestions to why is this happening
btw i also tried removing the while statement.

Answer 1

I think you mean the following while loop
while ( f < 0 || f > 3 )
{
printf("-press 1 to calculate the sum of a and b\n");
printf("-press 2 to calculate the difference between a and b\n");
printf("-press 3 to calculate the product of a and b\n");
scanf( "%d", &g );
f = g;
}
That is pay attention to 1) the condition of the while loop, 2) removed the return statement and 3) removed the character '\n' in scanf because it is redundant.
And in printf calls remove the operator & and use the conversion specifier %d.
if ( f == 1 ) {
printf(" A+B= %d\n", c) ;
}
else if ( f == 2) {
printf(" A-B= %d\n", d );
}
else if ( f == 3 ) {
printf(" A*B= %d\n",e );
}

Answer 2

I fixed you code and added some comments to it. Hope you can understand it! If you have questions, just comment them.
#include <stdio.h>
int main (void) {
// values can be given to ints by int a = value
int a = 19, b = 11, c, d, e, f = -1, g;
// '||' means OR, while '&&' means AND.
// A value can't be smaller then and bigger then 3 at the same time
while (f < 0 || f > 3) {
printf("-press 1 to calculate the sum of a and b\n");
printf("-press 2 to calculate the difference between a and b\n");
printf("-press 3 to calculate the product of a and b\n");
scanf("%d", &g); //you don't need \n for scanf
f=g;
// return stops the program. In this case, this is not what you want
}
// For printf you don't need &e, &f, etc.
// Instead of calculating a value and storing it in a different int,
// you can just print the outcome of the calculation.
if (f == 1) {
printf("A + B = %d\n", a + b);
} else if (f == 2) {
printf("A - B = %d\n", a - b);
} else if (f == 3) {
printf(" A * B = %d\n", a * b);
}
return 0;
}

Answer 3

Try this:
while (f<0 && f>3){
printf("-press 1 to calculate the sum of a and b\n");
printf("-press 2 to calculate the difference between a and b\n");
printf("-press 3 to calculate the product of a and b\n");
scanf("%d\n",&g);
f=g;
// remove the "return" from here;
}
Also change :
printf(" A+B= %ls\n",&c);
To:
printf(" A+B= %d\n",c);
You don't need the '&' operator to print values, if used it will print the address and not the resultant value.

Answer 4

I would like to add that you could replace your if/else if statements with the switch/case . To be more specific (after changing some small mistakes in your code %d instead %ls and so on) the section of your code should look like this:
// code above
switch(f) {
case 1:
printf(" A+B= %d\n", c) ;
break; // if you want the program to stop after doing this operation
case 2:
printf(" A-B= %d\n", d );
break;
case 3:
printf(" A*B= %d\n",e );
break;
// default: // if the input is not one of the above(1,2 or 3)
// printf(" Please give a valid option. \n");
// break;