CODEWITHSUNDEEP
cfunction
Fabrizio Mazzoni
Fabrizio Mazzoni

Reputation: 1909

Beginner C loop function

I just started venturing into C coming from PHP. I'm having trouble with the printf() function when its called from within another function:

#include <stdio.h>

void printloop (int valx) {
    int x;
    for (x = valx; x < 5; x++) {
        printf("Value of x is: %d\n", x);
    }
}

int main() {
    printloop(5);
    return 0;
}

The program will compile and run but there is no output on screen.

Upvotes: 0

Views: 172

Answers (5)

Daniel Ojeda
Daniel Ojeda

Reputation: 101

In your main, your are calling printloop and giving it the value 5. This means that you're making valx = 5 and when you say x = valx inside of your loop, you're setting x equal to 5 as well. When your loop gets to its conditional, it will never run because because x is equal to 5, not less than it.

Upvotes: 1

jrmullen
jrmullen

Reputation: 346

What's wrong here is that your logic is saying 5 < 5, which is false. Your for loop is not executing because when your printloop function is invoked at printloop(5); it is passing the integer value of 5.

void printloop (int valx) {
int x;
for (x = valx; x < 5; x++) {
    printf("Value of x is: %d\n", x);
}

Your printloop function is receiving a value of 5, setting x inside of your function to x = 5.

When it is time for your for loop to execute you will have

void printloop (int valx) {
    int x = 5;
    for (5 = valx; 5 < 5; 5++) {
        printf("Value of x is: %d\n", x);
    }
}

the for loop will see 5 < 5, which is false, and thus the loop will not execute.

I think what you want to say is 

#include <stdio.h>

void printloop (int valx) {
    int x;
    for (x = 0; x < valx; x++) {
        printf("Value of x is: %d\n", x);
    }
}

int main() {
    printloop(5);
    return 0;
}

Which will output:

Value of x is: 0
Value of x is: 1
Value of x is: 2
Value of x is: 3
Value of x is: 4

I hope this makes sense, and keep up the good work!

Upvotes: 5

klee8_
klee8_

Reputation: 187

It's because you passed 5 to the function printloop, and the for loop inside printloop will only run if the given argument is less than 5. You could either change that value or pass an integer smaller than 5.

Upvotes: 0

Daniel Kay
Daniel Kay

Reputation: 1

#include <stdio.h>

void printloop (int valx) {
int x;
    for (x = valx; x < 5; x++) {<--- **Your problem is here ( x = 5; x < 5; x++)
    printf("Value of x is: %d\n", x);
    }
}

int main() {
    printloop(5);
 return 0;
}

You are telling the loop while it is less than 5 (which it is not because of the value you gave to valx) thus it doesn't print anything because it doesn't go into the loop

Upvotes: 0

Crozin
Crozin

Reputation: 44376

When you invoke the printloop function with 5 the for loop essentially becomes

for (x = 5; x < 5; x++) {
    //...
}

And x < 5 will never be true.

I guess what you meant was something like 

for (x = 0; x < valx; x++) {
    //...
}

Upvotes: 6

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