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I'm having some issues trying to remove all whitespace from a string (for example: "a b c x") using a method that takes a double pointer argument. I have solved it using single pointers and using strings in c++, but I wanted to get this working as well for curiosity's sake :)

This is what I've tried:

void remove_whitespace(char** s)
{
    char **rd; // "read" pointer
    char **wr; // "write" pointer

    rd = wr = s; // initially, both read and write pointers point to beginning of string

    while ( *rd != 0 ) // while not at end of string
    {
    while ( isspace( **rd ) ) // while rd points to a whitespace character
        rd++;                  // advance rd
    *wr++ = *rd++;           // copy *rd to *wr, advance both pointers
    }
    *wr = 0;                   // terminate the result

    printf("\nsquished: %s\n", s);
}
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11
  • 6
    Why char** over char*? Commented Jan 13, 2023 at 16:58
  • 1
    Why are you using char ** for a single string? You don't reassign s so there's no need. Commented Jan 13, 2023 at 16:58
  • 1
    should be printf("\nsquished: %s\n", *s);, "%s" expects a char* Commented Jan 13, 2023 at 17:00
  • 1
    If you need to have char** s you can have char *rd, *wr; rd = wr = *s;. Commented Jan 13, 2023 at 17:03
  • 1
    Jonas Örnfelt, post how remove_whitespace() is called. Post a minimal reproducible example. Commented Jan 13, 2023 at 17:06

2 Answers 2

Reset to default
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All errors in this function stem from confusing the level of indirection. Using char** over char* increases the level of indirection which affects every usage of these variables.

Simply converting the function to use char* makes it much clearer:

void remove_whitespace(char* s)
{
    char *rd; // "read" pointer
    char *wr; // "write" pointer

    rd = wr = s; // initially, both read and write pointers point to beginning of string

    while ( *rd != 0 ) // while not at end of string
    {
        if (isspace( *rd ) ) // while rd points to a whitespace character
        {
            rd++;                  // advance rd
            continue;
        }
        *wr++ = *rd++;           // copy *rd to *wr, advance both pointers
    }
    *wr = 0;                   // terminate the result

    printf("\nsquished: %s\n", s);
}
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6 Comments

In the first version, *wr++; and *rd++; should be (*wr)++; and (*rd)++;.
'Here is a correct version while keeping char**' – have you tested? This looks pretty suspicious to me: *rd++; increments the pointee of rd, but due to rd = wr = s; this is the same pointee than the one of wr, so you increment *wr right with it, and actually *s, too!
You guys are correct. Fixed.
There is still one uncorrected *rd++;. Also, @Aconcagua's point about *s being modified still holds.
Probably best just to get rid of the first version. It's neither correct nor useful!
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  1. It is good if function returns something.
  2. Your local worker pointers do not have to be pointers to pointers
char **remove_whitespace(char **s)
{
    char *rd; // "read" pointer
    char *wr; // "write" pointer

    rd = wr = *s; 

    while ( *rd) // while not at end of string
    {
    while ( isspace( (unsigned char)*rd ) ) // while rd points to a whitespace character
        rd++;                  // advance rd
    *wr++ = *rd++;           // copy *rd to *wr, advance both pointers
    }
    *wr = 0;                   // terminate the result

    printf("\nsquished: %s\n", *s);
    return s;
}

Usage:

int main(void)
{
    char z[] = "asdas das as \r dfds \n dsa \t";
    char *p = z;

    remove_whitespace(&p);
}

https://godbolt.org/z/Ex337TE6n

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5 Comments

@JonasÖrnfelt because you do not pass pointer to pointer. READ THE WARNINGS IN YOUR CODE!!!! If you pass &z it will not give pointer to pointer only pointer to array and it is not thee same
'Your local worker pointers do not have to be pointers to pointers' – actually even: 'must not' – at least if initialised as in the question...
It is good if function returns something. Depends... In many cases convenient, but opens the question: Is the string returned a copy (i.e. does it have to be freed) or is the modification in place? Sure, documentation needs to reveal that anyway, but a void return type would automatically imply the latter (self-documenting code)...
Ah, okay. I was calling it with: char* test = "a b c "; remove_whitespace(&test);
You can't modify the string literal.

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