Why is this code producing an infinite loop?

Question

#include <Stdio.h>
#include <string.h>

int main(){
    char str[51];
    int k = 1;
    printf("Enter string\n");
    scanf("%s", &str);
    for(int i = 0; i < strlen(str); i++){
        while(str[k] != '\0')){
            if(str[i] == str[k]){
                printf("%c", str[i]);
                k++;
            }
        }
    }

    return 0;
}

It is simple C code that checks for duplicate characters in string and prints the characters. I am not understanding why it is producing an infinite loop. The inner while loop should stop when str[k] reaches the null terminator but the program continues infinitely.

Answer 1

Points to know

• You don't need to pass the address of the variable str to scanf()
• Don't use "%s", use "%<WIDTH>s", to avoid buffer-overflow
• Always check whether scanf() conversion was successful or not, by checking its return value
• Always use size_t to iterator over any array
• i < strlen(str), makes the loop's time complexity O(n³), instead of O(n²), which also isn't very good you should check whether str[i] != 0. But, many modern compilers of C will optimize it by the way.
• #include <Stdio.h> it is very wrong, stdio.h != Stdio.h
• Call to printf() can be optimized using puts() and putc() without any special formatting, here also modern compiler can optimize it
• while(str[k] != '\0')){ has a bracket (')')
• Initialize your variable str using {}, this will assign 0 to all the elements of str

---

Better Implementation

My implementation for this problem is that create a list of character (256 max) with 0 initialized, and then add 1 to ASCII value of the character (from str) in that list. After that print those character whose value was greater than 1.

• Time Complexity = O(n), where n is the length of the string

• Space Complexity = O(NO_OF_CHARACTERS), where NO_OF_CHARACTERS is 256

  ---

Final Code

#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

static void print_dup(const char *str)
{
    size_t *count = calloc(1 << CHAR_BIT, sizeof(size_t));
    for(size_t i = 0; str[i]; i++)
    {
        count[(unsigned char)str[i]]++;
    }
    for (size_t i = 0; i < (1 << CHAR_BIT); i++)
    {
        if(count[i] > 1)
        {
            printf("`%c`, count = %zu\n", i, count[i]);
        }
    }
    free(count);
}

int main(void) {
    char str[51] = {};
    puts("Enter string:");
    if (scanf("%50s", str) != 1)
    {
        perror("bad input");
        return EXIT_FAILURE;
    }
    print_dup(str);
    return EXIT_SUCCESS;
}

Answer 2

Read your code in English: You only increment variable k if character at index k is equal to character at index i. For any string that has different first two characters you will encounter infinite loop: char at index i==0 is not equal to char at index k==1, so k is not incremented and while(str[k]!=0) loops forever.