hi, my name is vic...i got one problem cannot fix....output for the number of times each of the tasks has been invoked...i run 3 times each of the tasks but the output always is this program has run 1 times.....

#include<stdio.h>
#include<stdlib.h>

void doTaskA(){
    printf("Start of Task A\n\n");
    int input_1, input_2, input_3;
    printf("Start of TASK A\n\n");
    printf("Please enter 3 integer.\n");
    scanf("%d\n%d\n%d",&input_1, &input_2, &input_3);
    if ((input_3>input_1) && (input_3>input_2))
        printf("/nThe biggest number is %d\n", input_3);
    else if ((input_2>input_1) && (input_2>input_3))
        printf("/nThe biggest number is %d\n", input_2);
    else 
        printf("/nThe biggest number is %d\n",input_1);

}



void doTaskB(){
int row, c, n, temp;

   printf("Start of Task B\n\n");
printf("Enter the number of rows in pyramid of stars you wish to see ");
   scanf("%d",&n);

   temp = n;

   for ( row = 1 ; row <= n ; row++ )
   {
      for ( c = 1 ; c < temp ; c++ )
         printf(" ");

      temp--;

      for ( c = 1 ; c <= 2*row - 1 ; c++ )
         printf("*");

      printf("\n");
   }

}



void doTaskC(){
 int n, c, k, space;
  printf("Start of Task c\n\n");

  scanf("%d", &n);

   space = 0;

   for ( k = n ; k >= 1 ; k-- )
   {
       for ( c = 1 ; c <= space ; c++ )
       printf(" ");

       space++;

       for ( c = 1 ; c <= k ; c++)
          printf("%d", c);

       printf("\n");
   }


}



void displayMenu(int &option) {



   printf("\nPlease choose one of the following :");
   printf("\n\n1. Task A\n");
   printf("2. Task B\n");
   printf("3. Task C\n");
   printf("4. Task D ( Exit This Program )\n");


   printf("\nPlease enter your selection :");
   scanf("%d",&option);

}

void howMany(){   

   static int count = 0;
   count++;


   printf("This program has run"  " %d "  "times\n\n\n",count);

   system("pause");

  }

 int main(){
    int option =0;
    bool exit = false;
    while(!exit) {
displayMenu(option);

        switch(option) {
            case 1:
                doTaskA();
                continue; 
            case 2:
                doTaskB();
                continue;
            case 3:
                doTaskC();
                continue;
            case 4:
                howMany();
                exit = true;
                break;
            default:
                printf("Incorrect selection. Try again.\n");
                continue;
}
}

    return 0;
}

Edited 3 Years Ago by Dani: Formatting fixed

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Comments
It's not hard to get the code into proper formatting.

OK, look at your main function:

int main(){
	int option =0;
	bool exit = false;
	while(!exit) {
		displayMenu(option);

		switch(option) {
		case 1:
			doTaskA();
			continue;
		case 2:
			doTaskB();
			continue;
		case 3:
			doTaskC();
			continue;
		case 4:
			howMany();
			exit = true;
			break;
		default:
			printf("Incorrect selection. Try again.\n");
			continue;
		}
	}

	return 0;
}

I'm not sure why you're using continue instead of break. The break would only exit out of the switch block and not the while loop.
So... after each of the tasks, you should be incrementing a counter. However, that counter is located here:

void howMany(){

	static int count = 0;
	count++;

	printf("This program has run" " %d " "times\n\n\n",count);

	system("pause");

}

HowMany is only getting called at the end of the program. Therein lies your problem.
My recommendation: yank the count variable declaration from howMany and make it global. Then, after each task is run, put count++; after each of their respective blocks (except number 4 and the default) in the switch.

This article has been dead for over six months. Start a new discussion instead.