Why would a character array be unchanged after a for-loop? - c

Question

I have built a function with the goal of taking text that is fed from elsewhere in the program and removing all whitespace and punctuation from it. I'm able to remove whitespace and punctuation, but the changes don't stay after they are made. For instance, I put the character array/string into a for-loop to remove whitespace and verify that the whitespace is removed by printing the current string to the screen. When I send the string through a loop to remove punctuation, though, it acts as though I did not remove whitespace from earlier. This is an example of what I'm talking about:
Example of output to screen
The function that I'm using is here.
//eliminates all punctuation, capital letters, and whitespace in plaintext
char *formatPlainText(char *plainText) {
int length = strlen(plainText);
//turn capital letters into lower case letters
for (int i = 0; i < length; i++)
plainText[i] = tolower(plainText[i]);
//remove whitespace
for (int i = 0; i < length; i++) {
if (plainText[i] == ' ')
plainText[i] = plainText[i++];
printf("%c", plainText[i]);
}
printf("\n\n");
//remove punctuation from text
for (int i = 0; i < length; i++) {
if (ispunct(plainText[i]))
plainText[i] = plainText[i++];
printf("%c", plainText[i]);
}
}
Any help as to why the text is unchanged after if exits the loop would be appreciated.

Answer 1

Those for loops are not necessary. Your function can be modified as follows and I commented where I made those changes:
char* formatPlainText(char *plainText)
{
char *dest = plainText; //dest to hold the modified version of plainText
while ( *plainText ) // as far as *plainText is not '\0'
{
int k = tolower(*plainText);
if( !ispunct(k) && k != ' ') // check each char for ' ' and any punctuation mark
*dest++ = tolower(*plainText); // place the lower case of *plainText to *dest and increment dest
plainText++;
}
*dest = '\0'; // This is important because in the while loop we escape it
return dest;
}
From main:
int main( void ){
char str[] = "Practice ????? &&!!! makes ??progress!!!!!";
char * res = formatPlainText(str);
printf("%s \n", str);
}

Answer 2

The code does convert the string to lower case, but the space and punctuation removal phases are broken: plainText[i] = plainText[i++]; has undefined behavior because you use i and modify it elsewhere in the same expression.
Furthermore, you do not return plainText from the function. Depending on how you use the function, this leads to undefined behavior if you store the return value to a pointer and later dereference it.
You can fix the problems by using 2 different index variables for reading and writing to the string when removing characters.
Note too that you should not use a length variable as the string length changes in the second and third phase. Texting for the null terminator is simpler.
Also note that tolower() and ispunct() and other functions from <ctype.h> are only defined for argument values in the range 0..UCHAR_MAX and the special negative value EOF. char arguments must be cast as (unsigned char) to avoid undefined behavior on negative char values on platforms where char is signed by default.
Here is a modified version:
#include <ctype.h>
//eliminate all punctuation, capital letters, and whitespace in plaintext
char *formatPlainText(char *plainText) {
size_t i, j;
//turn capital letters into lower case letters
for (i = 0; plainText[i] != '\0'; i++) {
plainText[i] = tolower((unsigned char)plainText[i]);
}
printf("lowercase: %s\n", plainText);
//remove whitespace
for (i = j = 0; plainText[i] != '\0'; i++) {
if (plainText[i] != ' ')
plainText[j++] = plainText[i];
}
plainText[j] = '\0';
printf("no white space: %s\n", plainText);
//remove punctuation from text
for (i = j = 0; plainText[i] != '\0'; i++) {
if (!ispunct((unsigned char)plainText[i]))
plainText[j++] = plainText[i];
}
plainText[j] = '\0';
printf("no punctuation: %s\n", plainText);
return plainText;
}