CODEWITHSUNDEEP
cfor-loopswitch-statementcs50
dannylee
dannylee

Reputation: 27

Repetition using switch statement inside a for loop

I am trying to use a loop to print out a repetitive song, "this old man" The first verse is: This old man, he played one He played knick-knack on my thumb This old man came rolling home

This song repeats to ten, varying the two terms in italicize one -> two++ and thumb -> another item such as shoe, knee, etc. Here is my code so far:

#include <cs50.h>
#include <stdio.h>
#include <string.h>

string change1 (int i);
int main (void)
{ 
    for (int i = 1; ; 1 < 11; i++)
    {
        printf ("This old man, he played "); 
        change1(i);
        printf("He played knick-knack on my %s\n\n", s1);
    }

    return 0;
 }

string change1(int i)
{
    string s1;

    switch(i)
    {
        case 1: 
        {
            printf("one\n");
            s1 = "thumb";
        }
            break;
        case 2:
        {
            printf("two\n");
            s1 = "shoe";
        }
            break;
        case 3:
        case 4:
        case 5:
        case 6:
        case 7:
        case 8:
        case 9:
        case 10:
        case 11:
            printf("ill add these cases later");
    }
}

This gives me an error message of: "control reaches end of non-void function"

I also got an undeclared variable s1 error but I declared it in the function.

Upvotes: 1

Views: 1115

Answers (4)

Keith Nicholas
Keith Nicholas

Reputation: 44298

You could simplify your program to an actual C program, rather than C++

int main (void)
{
    int i;
    char* items[] = {"thumb", "shoe", "", "", "", "", "", "", "", ""};
    char* numbers[] = {"one", "two", "three","four","five","six","seven","eight","nine","ten"};
    for (i = 0; i < 10; i++)
    {
        printf ("This old man, he played %s\n", numbers[i]);    
        printf("He played knick-knack on my %s\n\n", items[i]);
    }
  return 0
}

Upvotes: 2

Sergey Kalinichenko
Sergey Kalinichenko

Reputation: 726599

In C++ variables have scope. A variable is generally visible inside the curly braces where it is declared; outside these brackets the variable does not exist.

That is why you cannot use s1 from change1 inside the loop: you need to return a value (best choice in your situation), or use a variable that is in scope in both change1 and main.

printf ("This old man, he played ");
printf("He played knick-knack on my %s\n\n", change1(i));
...
string change1 (int i) {
    string s1;
    switch (i) {
        ...
    }
    return s1;
}

Note that you do not need a switch statement to implement change1: when the code is so uniform, you may be better off with an array:

const char *strings[] = {"thumb", "shoe", ...};

Upvotes: 0

Haji
Haji

Reputation: 2077

use return statement at the end of switch class

return s1;

Upvotes: -1

Barmar
Barmar

Reputation: 781058

change1 needs to return the string it decided on. And main has to assign the return value to a variable, because as originally written s1 is local to the change1 function.

#include <cs50.h>
#include <stdio.h>
#include <string.h>

string change1 (int i);
int main (void)
{ 
    for (int i = 1; ; 1 < 11; i++)
        {
            printf ("This old man, he played ");
            string s1 = change1(i);
            printf("He played knick-knack on my %s\n\n", s1);
        }
    return 0;
}

string change1 (int i)
{
    string s1;

    switch (i)
        {
        case 1: 
            {
                printf("one\n");
                s1 = "thumb";
            }
            break;
        case 2:
            {
                printf("two\n");
                s1 = "shoe";
            }
            break;
        case 3:
        case 4:
        case 5:
        case 6:
        case 7:
        case 8:
        case 9:
        case 10:
        case 11:
            printf("ill add these cases later");
        }
    return s1;
}

Upvotes: 0

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