print 1-50 as a block of 5 rows 10 columns using a function

[jimix]

o.k. so i can't program and this is child's play to most of your.
I am struggling with understanding functions. Here's what I have.#include <stdio.h>

    void row (int);
    int main (void)
 
{
    int x=41;
    int i;
 
 
    for (i=1; i<=x; i+=10)
        printf("%d",i);
    row (int j);
 
}
 
    void row(int j)
{    
    int j;
    for (j=1;j<=41;j++)
        printf("%d",j);
        return 0;
}

so i am trying to get
1 2 3 .... 10
11 12 13 .....20
21 22 23 .....30 up to 50.
I have to use a function.
I am trying to print the first column value,(i) pass it to the function,(row (int j)) add 1 to the value (i) (9 times) and then go back to main for my next value (i) wich will be 11 and so on.
Is this the correct approach? I've been at it for awhile now with no luck.

[Ganon11]

    void row (int);
    int main (void)
 
{
    int x=41;
    int i;
 
 
    for (i=1; i<=x; i+=10)
        printf("%d",i);
    row (int j);
 
}
 
    void row(int j)
{    
    int j;
    for (j=1;j<=41;j++)
        printf("%d",j);
        return 0;
}

This code is wrong on several counts. I'll try and explain them all.

1) Why do you have printf statements in main if you're going to use your row function to print the numbers in nice rows?
2) When you call rows, you say int j - which (might?) create a new variable j that has been uninitialized.
3) In rows, you are passed j, but then you declare a new variable also named j, so the argument j is overwritten (or your code won't even compile). How are you going to print the value when you've hidden it from the function?
4) row is a void function, but you return 0.
5) main is an int function, but you don't return anything.

I have a feeling you didn't try and compile this code - or if you did, you saw all the warnings and didn't try to fix them. Get this code to compile first, then worry about the formatting - which, by the way, is going to be more difficult than just printing numbers in a for loop.