Implicit Declaration of Function in C on CodeBlocks

Question

Hello I am practicing my knowledge in C language I am trying to make a simple calculator but I encountered this warning Implicit Declaration of Function but the function that I called has been executed. I tried to fix it with this void start(); but the function did not execute.

Successful executed the function start(); but have a implicit warning:

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

void addition()
{
    int vala, valb, resu;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("ADDITION\n");
    printf("\n");
    printf("Enter the first value of addend: ");
    scanf("%d", &vala);
    printf("Enter the second value of addend: ");
    scanf("%d", &valb);
    resu=vala+valb;
    printf("The sum of %d and %d is: %d\n", vala, valb, resu);
    printf("PRESS [ANY KEY] TO CONTINUE...");
    getch();

    start(); \\THIS CODE
}

void start()
{
    char ope;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("What operation will be used:");
    scanf("%s", &ope);

    if (ope == 'a')
    {
        addition();
    }
    else if (ope == 'b')
    {
        printf("bbbbbbbbbbbb\n");
    }
    else
    {
        printf("ccccccccccc\n");
    }

}

int main()
{
    int choices;
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("choose an option:");
    scanf("%d", &choices);

    if (choices == 1)
    {
        start();
    }

    getch();
    return 0;
}

Failed to execute the function void start(); start but no implicit warning:

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

void addition()
{
    int vala, valb, resu;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("ADDITION\n");
    printf("\n");
    printf("Enter the first value of addend: ");
    scanf("%d", &vala);
    printf("Enter the second value of addend: ");
    scanf("%d", &valb);
    resu=vala+valb;
    printf("The sum of %d and %d is: %d\n", vala, valb, resu);
    printf("PRESS [ANY KEY] TO CONTINUE...");
    getch();

    void start(); \\THIS CODE
}

void start()
{
    char ope;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("What operation will be used:");
    scanf("%s", &ope);

    if (ope == 'a')
    {
        addition();
    }
    else if (ope == 'b')
    {
        printf("bbbbbbbbbbbb\n");
    }
    else
    {
        printf("ccccccccccc\n");
    }

}

int main()
{
    int choices;
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("choose an option:");
    scanf("%d", &choices);

    if (choices == 1)
    {
        start();
    }

    getch();
    return 0;
}

Answer 1

start() is declare after addition() , but in addition() you call start() , so the compiler don't know what start() is. Also, in start() you also call addition() , so the best way to reslove this is using forward declaration:

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

void start(void); /* forward declaration */
void addition(void); /* forward declaration */

void addition(void)
{
    int vala, valb, resu;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("ADDITION\n");
    printf("\n");
    printf("Enter the first value of addend: ");
    scanf("%d", &vala);
    printf("Enter the second value of addend: ");
    scanf("%d", &valb);
    resu=vala+valb;
    printf("The sum of %d and %d is: %d\n", vala, valb, resu);
    printf("PRESS [ANY KEY] TO CONTINUE...");
    getch();

    start(); 
}

void start(void)
{
    char ope;
    system("cls");
    printf("SiMPLE CALCULATOR 1.0a\n");
    printf("What operation will be used:");
    scanf("%s", &ope);

    if (ope == 'a')
    {
        addition();
    }
    else if (ope == 'b')
    {
        printf("bbbbbbbbbbbb\n");
    }
    else
    {
        printf("ccccccccccc\n");
    }

}

Answer 2

It is not advisable to take measures which allow you to call start from addition , leading to a potentially infinite recursion, only to fake a loop. Better drop the start(); call in addition and replace the scanf("%s", &ope); with while (scanf(" %c", &ope) > 0) .