scanf doesn't work properly for third time - c

Question

I'm trying to write a simple C program. The input format is like this:
Enter a: <a>
I want to input a, b, c and d but the scanf doesn't work for c and d and the program crashes on that point.
#include <stdio.h>
#include <math.h>
int main(){
int a, b, c, d;
char ch;
printf("Enter a:");
scanf("%c%d%c", &ch, &a, &ch);
printf("Enter b:");
scanf("%c%d%c", &ch, &b, &ch);
printf("Enter c:");
scanf("%c%d%c", &ch, &c, &ch);
printf("Enter d:");
scanf("%c%d%c", &ch, &d, &ch);
double ans;
ans=(a*b/c + c*d/a)/(b*a/c + d*b/a);
ans=fabs(ans);
printf("Formula 1 result = <%lf>", ans);
return 0;
}

Answer 1

There are 3 format conversion specifiers which do not skip leading white space. They are %c (which you're using), %[…] scansets, and %n.
When you type:
Enter a:<1>
Enter b:<2>
The first < satisfies the first %c; the 1 matches the %d, the > matches the second %c, so all is fine so far.
The second lot of input, though, reads the newline from the first line into the first %c, then fails to convert the < to an integer and stops, leaving the < for the next input to process.
The Enter c: prompt is given, but the <2> is processed because it is in the correct format, so the Enter d: prompt is given.
There are multiple ways to fix this, including:
Insert a blank before the first %c in the format strings. This will skip white space, including newlines, and give you the expected behaviour. It is probably the simplest fix.
Read lines and use sscanf() to process them. This is generally the best solution. (Use fgets() to read the line.)
In either case, check that the scanf() family function returns the correct value (3). And use two variables int c1, c2; to read the characters and validate that the user typed what you require.
A bare minimum fixup is:
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
int main(void)
{
int a, b, c, d;
char c1, c2;
printf("Enter a: ");
if (scanf(" %c%d%c", &c1, &a, &c2) != 3 || c1 != '<' || c2 != '>')
fprintf(stderr, "Bogus input!\n"), exit(1);
printf("Enter b: ");
if (scanf(" %c%d%c", &c1, &b, &c2) != 3 || c1 != '<' || c2 != '>')
fprintf(stderr, "Bogus input!\n"), exit(1);
printf("Enter c: ");
if (scanf(" %c%d%c", &c1, &c, &c2) != 3 || c1 != '<' || c2 != '>')
fprintf(stderr, "Bogus input!\n"), exit(1);
printf("Enter d: ");
if (scanf(" %c%d%c", &c1, &d, &c2) != 3 || c1 != '<' || c2 != '>')
fprintf(stderr, "Bogus input!\n"), exit(1);
double ans;
ans=(a*b/c + c*d/a)/(b*a/c + d*b/a);
ans=fabs(ans);
printf("Formula 1 result = <%lf>\n", ans);
return 0;
}
I'm being lazy with the comma operators after the if statement. It would be better, though, to use a function to manage the inputs:
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
static void get_number(int *var, const char *name)
{
char c1, c2;
printf("Enter %s: ", name);
if (scanf(" %c%d%c", &c1, var, &c2) != 3 || c1 != '<' || c2 != '>')
{
fprintf(stderr, "Bogus input!\n");
exit(1);
}
}
int main(void)
{
int a, b, c, d;
get_number(&a, "a");
get_number(&b, "b");
get_number(&c, "c");
get_number(&d, "d");
double ans;
ans=(a*b/c + c*d/a)/(b*a/c + d*b/a);
ans=fabs(ans);
printf("Formula 1 result = <%lf>\n", ans);
return 0;
}
And better still to use:
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
static int get_number(const char *name)
{
char buffer[4096];
printf("Enter %s: ", name);
if (fgets(buffer, sizeof(buffer), stdin) == 0)
{
fprintf(stderr, "EOF or error on input\n");
exit(1);
}
char c1, c2;
int val;
if (sscanf(buffer, " %c%d%c", &c1, &val, &c2) != 3 || c1 != '<' || c2 != '>')
{
/* buffer contains a newline unless you typed more than 4095 characters before the newline */
fprintf(stderr, "Bogus input (does not match <number>): %s", buffer);
exit(1);
}
return val;
}
int main(void)
{
int a = get_number("a");
int b = get_number("b");
int c = get_number("c");
int d = get_number("d");
double ans = fabs((a*b/c + c*d/a) / (b*a/c + d*b/a));
printf("Inputs: a = %d, b = %d, c = %d, d = %d\n", a, b, c, d);
printf("Formula 1 result = <%lf>\n", ans);
return 0;
}
Note how the code uses fgets() plus sscanf() which means that the error message can tell the user what they entered that was not what was expected. It also uses the function to return the value, leading to the simpler code shown. And the output includes the input values for inspection. That may not be needed for your submission system, but it surely helps you while you're developing the code.
You can work on the upgrade to process double values as inputs; it isn't hard.

Answer 2

double ans;
ans=(a*b/c + c*d/a)/(b*a/c + d*b/a);
The C language doesn't take into account what you do with a result when it determines its type. So the fact that you assign the results of this operation to a double don't have any effect on the way the result is computed, it just converts the integer result to a double. When you divide integers by integers, you get integer division.
You could do something like this:
ans=(a*1.0*b/c + c*1.0*d/a)/(b*1.0*a/c + d*1.0*b/a);
The multiplication by a non-integer forces non-integer multiplication, addition, and division to be used.