Our activity requires to input 4 numbers, positive or negative and only add the negative numbers.

Example:

-30.22
10.50
-2.2
-1.8

Result is -34.22 which sums up the negative numbers only.

We have not discussed the loop or array at this moment. Is it possible to solve in a mathematical equation? I can’t seem to find any answers if I try to code it.

Can it be solved using if-else-elseif statements?

CodePudding user response：

Read the input into four variables, then use four if statements to add them to the total when they're negative.

float a, b, c, d;
scanf("%f %f %f %f", &a, &b, &c, &d);
float total = 0;
if (a < 0) {
    total  = a;
}
if (b < 0) {
    total  = b;
}
if (c < 0) {
    total  = c;
}
if (d < 0) {
    total  = d;
}
printf("%.2f\n", total);

You can also use the conditional operator.

float total = (a < 0 ? a : 0)   (b < 0 ? b : 0)   (c < 0 ? c : 0)   (d < 0 ? d : 0);

CodePudding user response：

These assignments some teachers invent are really pathetic.

Maybe you want this:

#include <stdio.h>

int main(void)
{
  float a, b, c, d;
  scanf("%f %f %f %f", &a, &b, &c, &d);
  float total = (a < 0 ? a : 0)   (b < 0 ? b : 0)   (c < 0 ? c : 0)   (d < 0 ? d : 0);
  printf("%f", total);
}

No loops, no if/else.

But be aware that this is really terrible code.

CodePudding user response：

A solution that doesn't use neither if nor while (and co.)

int main(void) {
  float n1 = 0, n2 = 0, n3 = 0, n4 = 0;

  printf("Insert first number: ");
  scanf("-%f", &n1);
  scanf("%*[^\n]");

  printf("Insert second number: ");
  scanf(" -%f", &n2);
  scanf("%*[^\n]");

  printf("Insert third number: ");
  scanf(" -%f", &n3);
  scanf("%*[^\n]");

  printf("Insert fourth number: ");
  scanf(" -%f", &n4);
  scanf("%*[^\n]");

  float sum = -n1 - n2 - n3 - n4;

  printf("Sum of all negative numbers: %f\n", sum);

  return 0;
}

Explanation

scanf("-%f", &n1) will have two possible behaviors:

• in case of positive numbers (e.g.: 3.14 or 1.2), it will fail to parse the input (because it isn't able to match the first - symbol), so variable n1 stays 0;
• in case of negative numbers (e.g.: -5 or -3.1), it will match the minus symbol and put the rest of the number into n1 (so n1 is either 5 or 3.1, for instance).

Basically, for all four numbers, if the user input is positive, n is 0; if the input is negative, n is the absolute value of the number (2.0 instead of -2.0 etc).

scanf("%*[^\n]") will clear the input buffer after (possible) positive numbers (because scanf() won't match positive numbers).

float sum = -n1 - n2 - n3 - n4 is a (weird) "sum", but necessary since all four numbers are either 0 or the negation of a negative number.