在 C 中操作动态分配的二维字符数组

Question

I'm having trouble with trying to manipulate 2d dynamic arrays in C. What I want to do is to store a char string in every row of the the 2d array then perform a check to see if the string contains a certain character, if so remove all occurrences then shift over the empty positions. What's actually happening is I get an exit status 1 .

More about the problem, for example if I have

Enter string 1: testing
Enter string 2: apple
Enter string 3: banana

I would want the output to become

What letter? a // ask what character to search for and remove all occurences
testing
pple
bnn

Here is my full code:

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


  void removeOccurences2(char** letters, int strs, int size, char letter){
    // Get size of array
    // Shift amount says how many of the letter that we have removed so far.
    int shiftAmt = 0;
    // Shift array says how much we should shift each element at the end
    int shiftArray[strs][size];

    // The first loop to remove letters and put things the shift amount in the array
    int i,j;
    for(i=0;i < strs; i++){
        for(j = 0; j < size - 1; j++) {

              if (letters[i][j] == '\0'){
                  break;
              }

              else {
              // If the letter matches

              if(letter == letters[i][j]){

              // Set to null terminator
              letters[i][j] = '\0';
              // Increase Shift amount
              shiftAmt++;
              // Set shift amount for this position to be 0
              shiftArray[i][j] = 0;
              }else{
              // Set the shift amount for this letter to be equal to the current shift amount
              shiftArray[i][j] = shiftAmt;
              }
              }
        }
      }  

    // Loop back through and shift each index the required amount
    for(i = 0; i < strs; i++){
        for(j = 0; j < size - 1; j++) {
          // If the shift amount for this index is 0 don't do anything


          if(shiftArray[i][j] == 0) continue;
          // Otherwise swap
          letters[i][j - shiftArray[i][j]] = letters[i][j];
          letters[i][j] = '\0';

        }

        //now print the new string
        printf("%s", letters[i]);
    }
    return;
  }

  int main() {
      int strs;
      char** array2;
      int size;
      int cnt;
      int c;
      char letter;
      printf("How many strings do you want to enter?\n");
      scanf("%d", &strs);
      printf("What is the max size of the strings?\n");
      scanf("%d", &size);
      array2 = malloc(sizeof(char*)*strs);
      cnt = 0;
      while (cnt < strs) {
          c = 0;
          printf("Enter string    %d:\n", cnt + 1);
          array2[cnt] = malloc(sizeof(char)*size);
          scanf("%s", array2[cnt]);
          cnt += 1;
      }

        printf("What letter?\n");
      scanf(" %c", &letter);
      removeOccurences2(array2,strs,size,letter);

  }

Thanks in advance!

Answer 1

You can remove letters from a string in place, because you can only shorten the string.

The code could simply be:

void removeOccurences2(char** letters, int strs, int size, char letter){
    int i,j,k;
    // loop over the array of strings
    for(i=0;i < strs; i++){
        // loop per string
        for(j = 0, k=0; j < size; j++) {
              // stop on the first null character
              if (letters[i][j] == '\0'){
                  letters[i][k] = 0;
                  break;
              }
              // If the letter does not match, keep the letter
              if(letter != letters[i][j]){
                  letters[i][k++] = letters[i][j];
              }
        }
        //now print the new string
        printf("%s\n", letters[i]);
    }
    return;
  }

But you should free all the allocated arrays before returning to environment, and explicitely return 0 at the end of main .

Answer 2

"...if the string contains a certain character, if so remove all occurrences then shift over the empty positions."

The original string can be edited in place by incrementing two pointers initially containing the same content. The following illustrates.:

void remove_all_chars(char* str, char c) 
{
    char *pr = str://pointer read
    char *pw = str;//pointer write
    while(*pr) 
    {
        *pw = *pr++;     
        pw += (*pw != c);//increment pw only if current position == c
    }
    *pw = '\0';//terminate to mark last position of modified string
}

This is the cleanest, simplest form I have seen for doing this task. Credit goes to this answer .

Answer 3

Well, there are several issues on your program, basically you are getting segmentation fault error because you are accessing invalid memory which isn't allocated by your program. Here are some issues I found:

1. shiftAmt isn't reset after processing/checking each string which lead to incorrect value of shiftArray .
2. Values of shiftArray only set as expected for length of string but after that (values from from length of each string to size ) are random numbers.
3. The logic to delete occurrence character is incorrect - you need to shift the whole string after the occurrence character to the left not just manipulating a single character like what you are doing.

1 & 2 cause the segmentation fault error (crash the program) because it causes this line letters[i][j - shiftArray[i][j]] = letters[i][j]; access to unexpected memory. You can take a look at my edited version of your removeOccurences2 method for reference:

int removeOccurences2(char* string, char letter) {
    if(!string) return -1;
    int i = 0;

    while (*(string+i) != '\0') {
        if (*(string+i) == letter) {
            memmove(string + i, string + i + 1, strlen(string + i + 1));
            string[strlen(string) - 1] = '\0'; // delete last character
        }
        i++;
    }
    return 0;
}

It's just an example and there is still some flaw in its logics waiting for you to complete. Hint: try the case: "bananaaaa123"

Happy coding!