为什么for循环在c中无限进行

Question

my question is when I run the code the loop goes on infinitely. the expected output is to print hi 5 times then go to a new line and print 4 times the 3 and so on I cant understand what is causing the loop to go infinite.

#include <stdio.h>
int star(int);
int main()
{
    printf("enter number\n");
    int a;
    scanf("%d", &a);
    star(a);
    return 0;
}

int star(int m)
{

    int n;
    for (n = 0; n < m; n++)
    {
        printf("hi");
    }
    printf("\n");

    star(m - 1);

    return 0;
}

Answer 1

you forgot to check if m > 0 somewhere in your function. In this case, the loop goes infinetely, because it doesn´t have a way to get out of recursion.

Simple fix:

#include <stdio.h>
int star(int);
int main()
{
    printf("enter number\n");
    int a;
    scanf("%d", &a);
    star(a);
    return 0;
}

int star(int m)
{

    int n;
    for (n = 0; n < m; n++)
    {
        printf("hi");
    }
    printf("\n");
    if (m > 0){
        star(m - 1);
    }
    return 0;
}

Answer 2

For starters the return type of the function star is useless and does not make a sense. Also there is no great sense to declare the parameter as having the signed integer type int . It would be more logically correct to declare the parameter as having at least the unsigned integer type unsigned int .

So the function should be declared like

void star( unsigned int n );

The function calls infinitely itself recursively because such a call is unconditional and does not depend on the value of the parameter m .

int star(int m)
{
    //...
    star(m - 1);
   
    return 0;
}

The function can be defined the following way as it is shown in the demonstrative program below.

#include <stdio.h>

void star( unsigned int );

int main( void )
{
    printf( "Enter number: " );
    unsigned int n;

    if ( scanf( "%u", &n) == 1 ) star( n );

    return 0;
}

void star( unsigned int n )
{
    if ( n )
    {
        for ( unsigned int i = 0; i < n; i++)
        {
            printf( "hi" );
        }

        putchar( '\n' );

        if ( --n ) star( n );
    }
}

Pay attention to that if the function was called initially with the argument equal to 0 then the function should output nothing.