C code doesn't work with but no errors

Question

I am writing the following code to reverse the string, without reversing the words. For e.g. I/P: This is sparta O/P: sparta is This

Here's the code:

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

void main()
{   
    char a[50],b[50];
    int i,j,k,l;
    printf("Enter a string: ");
    scanf("%[^\n]s",a);
    l=strlen(a);
    i=0;j=l-1;
ae: 
    for(i;;i++)
        if(a[i]==' ' || a[i]=='\0')
            break;
    k=i;
    for(j;;j--,k--)
    {
        if(a[k]==' ' || k==0)
            break;
        else
            b[j]=a[k-1];
    }
    if(strlen(b)!=l) goto ae;
    else 
        printf("%s",b);
    getch();
}   

This code runs wihout errors but after accepting the string, it doesn't go further at all. The terminal gets kinda hanged. I don't know whats the problem. Please help.

Answer 1

I debugged your program and the problem i found is that your program is entering into a never ending loop.

for(i;;i++)
    if(a[i]==' ' || a[i]=='\0')
        break;

the above loop breaks at i=4 when a[i] was equal to ' '. Later after goto is executed,i is still 4 and when

if(a[i]==' ' || a[i]=='\0')
        break;

is evaluated i remains 4 and therefore again breaks from the loop.Thus leading to a endless loop.

Answer 2

This is an infinite loop if the entered string contains a space.

1. Iterate over the string a until a[i] == ' '.
2. k = i;
3. Immediately break out of the second for loop because a[i] == ' ' and k == i implies a[k] == ' '.
4. b is uninitialized but strlen(b) != l will be true with high probability and therefore the branch goto ae; will be taken and the infinite loop is closed.

Answer 3

The format string in scanf("%[^\n]s",a); is made up of two parts: "%[^\n]" and "s".

The first part expects a sequence of characters different than '\n' and writes them in the corresponding address.

The second part expects a literal 's' and discards it.

With the input "This is sparta\n" the scanf() isn't terminated. It is still looking for an 's'.

Answer 4

You cannot safely use strlen(b), since the characters in b are not initialized to 0.

So to begin with, change b[50] to b[50] = {0}.