Question

CAN ANYONE HELP ME PLEASE! THUMBS UP WILL BE GIVEN!! THANKS
2.1 Use a text editor to enter the program on textbook page 47 (the code is given here for your convenience). Use GNU GCC to compile, debug (find and fix any syntax errors), and execute the program and fix any semantic errors, for example, by adding "break" statements in the required places and by fixing incorrect characters copied into the programming environment, if any, and by type-changing to print a floating point number. The code is supposed to perform one math operation in each switch-case. The ‘ch' variable is assigned a new math operator before each switch-case. Submit the corrected program as hw01q2.c [8 points] /* This C program demonstrates the switch statement without using breaks. */ #include <stdio.h> main) char ch int a 10, b20; double f; printf("ch-%c\n", ch); switch (ch) ( case '+': f- a + b; printf("f-%d\n", case '-': f- a - b; printf("f-%d\n", case '*': f- a * b; printf("f-%d\n", case '/': f- a / b; printf("f-%d\n", default: printf("invalid operator\n"); f); f); f); f); printf("ch-%c\n", ch); switch (ch) ( case '+': f- a + b; printf("f-%d\n", case '-': f- a - b; printf("f-%d\n", case '*': f- a * b; printf("f-%d\n", case '/': f- a / b; printf("f-%d\n", default: printf("invalid operator\n"); f); f); f); f); printf("ch-%c\n", ch); switch (ch) ( case '+': f- a + b; printf("f-%d\n", case '-': f- a - b; printf("f-%d\n", f); f
$case : f - a * b; printf(f-%d\n, case ./. : f -a / b; printf(f-%d\n, default: printf(invalid operator\n); f); f); ch printf(ch-%c\n, ch); switch (ch) case +: f a + b; printf( f-%d\n, case -: f a - b; printf(f-%d\n, case : f - a * b; printf(f-%d\n, case ./: f -a / b; printf(f-%d\n, default: printf(invalid operator\n); f); f); f); f); printf(ch-%c\n, ch); switch (ch) { case +: f a + b; printf(f-%d\n, case -: f a - b; printf(f-%d\n, case : f a * b; printf(f-%d\n, case /. : f a / b; printf(f-%d\n, default: printf(invalid operator\n); f); f); f); f);$

Answer 1

/* I think you want to change the provided code so that it prints only one math operation per switch statement, if this isn't the case do let me know in the comment section. */

#include <stdio.h>

int main() {
   char ch = '+';
   int a = 10, b = 20;
   double f;
   switch (ch) {
       case '+': f = a + b; printf("f = %f ", (float)f); break;
       case '-': f = a - b; printf("f = %f ", (float)f); break;
       case '*': f = a * b; printf("f = %f ", (float)f); break;
       case '/': f = a / b; printf("f = %f ", (float)f); break;
       default: printf("invalid operator ");
   }
   ch = '-';
   printf("ch = %c ", ch);
   switch (ch) {
       case '+': f = a + b; printf("f = %f ", (float)f); break;
       case '-': f = a - b; printf("f = %f ", (float)f); break;
       case '*': f = a * b; printf("f = %f ", (float)f); break;
       case '/': f = a / b; printf("f = %f ", (float)f); break;
       default: printf("invalid operator ");
   }
   ch = '*';
   printf("ch = %c ", ch);
   switch (ch) {
       case '+': f = a + b; printf("f = %f ", (float)f); break;
       case '-': f = a - b; printf("f = %f ", (float)f); break;
       case '*': f = a * b; printf("f = %f ", (float)f); break;
       case '/': f = a / b; printf("f = %f ", (float)f); break;
       default: printf("invalid operator ");
   }
   ch = '/';
   printf("ch = %c ", ch);
   switch (ch) {
       case '+': f = a + b; printf("f = %f ", (float)f); break;
       case '-': f = a - b; printf("f = %f ", (float)f); break;
       case '*': f = a * b; printf("f = %f ", (float)f); break;
       case '/': f = a / b; printf("f = %f ", (float)f); break;
       default: printf("invalid operator ");
   }
   ch = '%';
   printf("ch = %c ", ch);
   switch (ch) {
       case '+': f = a + b; printf("f = %f ", (float)f); break;
       case '-': f = a - b; printf("f = %f ", (float)f); break;
       case '*': f = a * b; printf("f = %f ", (float)f); break;
       case '/': f = a / b; printf("f = %f ", (float)f); break;
       default: printf("invalid operator ");
   }

   return 0;
}

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OUTPUT