C : trying to remove spaces from a string without using any string library functions

Question

So I am trying to figure out a way to print a string without spaces and it is not working for some reason, whenever I enter in a string to the program it ends out printing nothing.

#include <stdio.h>

char *removeSpaces(char *inString);

int main() {
    char str1[50];
    printf("Enter a string : ");
    scanf("%s", &str1);
    removeSpaces(str1);
}

char *removeSpaces(char *inString) {
    char str2[50];
    int j = 0;

    for (int i = 0; i < 50; i++) {
        if (inString[i] != ' ') {
            str2[j] = inString[i];
            j++;
        }
    }

    for (int i = 0; i < 50; i++) {
        printf("%s", str2[i]);
    }
}

Do not use `&` with arrays in scanf ==> `scanf("%49s", str1);` — pmg, Apr 09 '20 at 14:32
In `removeSpaces()` print characters with `printf("%c", str2[i]);` — pmg, Apr 09 '20 at 14:34
`%s` in `scanf()` will look for the next whitespace character and ends the input. Use `%[^\n]`. — Ruks, Apr 09 '20 at 14:48
@pmg hey so i tried printing it, it shows the first segment fine but when it reaches the space it prints garbled text — dankcodez, Apr 09 '20 at 14:49
@Ruks oh cheers man it worked, only problem is it shows a bunch of garbled text beside it — dankcodez, Apr 09 '20 at 14:51
Change the printing to `for (int i = 0; str2[i] != '\0'; i++) printf("%c", str2[i]);` OR simply get rid of the for loop and just do `printf("%s", str2);` — Eraklon, Apr 09 '20 at 14:56
@dankcodez The "garbled" text is due to undefined behavior because your array is uninitialized. Do this: `char str1[50] = {0};` for both the arrays. — Ruks, Apr 09 '20 at 14:59
@dankcodez Good reference to read more about that here: https://stackoverflow.com/questions/39431924/what-does-n-mean-in-c — PatrickOfThings, Apr 09 '20 at 15:31

ryyker · Answer 1 · 2020-04-09T17:50:35.257

You can also use pointers to walk memory locations testing each position for unwanted char value, Here is your function modified to use that method:

char *removeSpaces(char *inString);// your original prototype
char *remove_char(char *inString, char c);//generalized  

int main(void) {
    //use suggestions in other answer for correct user input
    const char str1[] = {"this is a string with spaces"};//for simple illustration

    printf("%s\n", removeSpaces(str1));
    printf("%s\n", remove_char(str1, ' '));
    return 0;
}

char *removeSpaces(char *inString)// your original prototype
{
    if(!inString) return (char *)"bad input";
    char *from; // "read" pointer
    char *to; // "write" pointer

    from = to = inString; // init both read and write pointers to original string

    while (*from) 
    { 
        if( (*from != ' ') && (*from != '\t') && (*from != '\n'))
        { 
            *to++ = *from; 
        } 
        from++; 
    } 
    *to = 0; 

    return inString;// add a return statement to return the char *
}

//optionally, create your function to remove any unwanted character
char *remove_char(char *inString, char c)//generalized  
{
    char *from; // "read" pointer
    char *to; // "write" pointer

    from = to = inString; // init both read and write pointers to original string

    while (*from) 
    { 
        if (*from != c) 
        { 
            *to++ = *from; 
        } 
        from++; 
    }  
   *to = 0; 

    return inString;// add a return statement to return the char *
}

chux - Reinstate Monica · Accepted Answer · 2020-04-09T15:57:48.253

Wrong input approach

Below will not scan into str1 anything with a space.

// bad
char str1[50];
scanf("%s", &str1);

Instead, use fgets() to read a line and form a string.

char str1[50];
if (fgets(str1, sizeof str, stdin)) {
  // success!

Lop off the potential trailing '\n' if desired.

  str1[strcspn(str1, "\n")] = '\0';

Reads past end of string

The loops reads to the size of the array. It should loop to the null chracter.

// for (int i = 0; i < 50; i++)
for (int i = 0; inString[i]; i++)

Missing \0

The string formation in str2[] is incomplete as it lacks a null chracter '\0'.

str2[j] = '\0'; // add after the loop

Warnings not fully enabled

Below should warn about mis-match of "%s" with str2[i].

for (int i = 0; i < 50; i++) {
    printf("%s", str2[i]);
}

Instead, without a loop.

 printf("%s", str2);

This is the biggest lesson to learn here. By fully enabling warnings, the compiler provides rapid feedback that something is wrong or questionable; faster than Stackoverflow.

Missing return

char *removeSpaces(char *inString) is expected to return a char *, yet code lacks a return something;.

C : trying to remove spaces from a string without using any string library functions

2 Answers2