Trying to fix - segmentation fault in C

Question

I'm executing a method to replace all spaces in a string with "%20". One may assume that the string has sufficient space at the end to hold the additional characters, and that one are given the "true" length of the string.

EXAMPLE:

Input: "Mr John Smith ", 13

Output: "Mr%20John%20Smith"

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

void URL(char str[], int length)
{

    int spacecount = 0, index;

    for (int i = 0; i < length; i++)
    {
        if (str[i] == ' ')
        {
            spacecount++;
        }
    }
    index = length + spacecount * 2;
    if (length < strlen(str))
        str[length] = '\0';

    for (int j = length - 1; j >= 0; j--)
    {
        if (str[j] == ' ')
        {
            str[index - 1] = '0';
            str[index - 2] = '2';
            str[index - 3] = '%';
            index = index - 3;
        }
        else
        {
            str[index - 1] = str[j];
            index--;
        }
    }
    printf("%s", str);
}

int main()
{
    char str[100];
    int len;
    printf("Enter the string : ");
    scanf("%s", str);
    printf("Enter the length : ");
    scanf("%d", &len);
    URL(str, len);
}

On executing the above code using gcc compiler, I'm getting the segmentation fault. I understand, what is segmentation fault. I want to fix it in this program.

Answer 1

Besides using fgets, as proposed by @4386427, you can also use scanf with the specifier %[^\n] (matches anything that isn't a \n) to read an entire line, like this:

scanf(" %[^\n]", str);

And for safety, when dealing with strings you should always limit the number of characters that can be read so that it doesn't exceed the length of the array str, by doing:

scanf(" %99[^\n]", str);

So your main could be:

int main()
{
    char str[100];
    int len;
    printf("Enter the string : ");
    scanf(" %99[^\n]", str);
    for (int c=getchar(); c!='\n' && c!=EOF; c=getchar());
    len = strlen(str);
    URL(str, len);
}

Answer 2

How does

scanf("%s", str);

work?

What will str contain if your input is: Mr John Smith

It's not doing what you think! The answer is that it will contain Mr and thereby you don't have a string of length 13.

Use fgets instead but remember to remove the newline.

The complete program could be:

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

void URL(char str[], int length)
{
    printf("input: |%s|\n", str);
    int spacecount = 0, index;

    for (int i = 0; i < length; i++)
    {
        if (str[i] == ' ')
        {
            spacecount++;
        }
    }
    index = length + spacecount * 2;
    str[index] = '\0';
    for (int j = length - 1; j >= 0; j--)
    {
        if (str[j] == ' ')
        {
            str[index - 1] = '0';
            str[index - 2] = '2';
            str[index - 3] = '%';
            index = index - 3;
        }
        else
        {
            str[index - 1] = str[j];
            index--;
        }
    }
    printf("output: |%s|\n", str);
}

int main()
{
    char str[100];
    int len;
    printf("Enter the string : ");
    fgets(str, 100, stdin);
    len = strlen(str);
    if (len > 0 && str[len-1] == '\n') str[--len] = '\0';
    URL(str, len);
}

Input:

Mr John Smith     

note: With 5 extra spaces after Smith

Output:

input: |Mr John Smith     |
output: |Mr%20John%20Smith%20%20%20%20%20|