help with the "transpose matrix" question

[queen crest]

I tried to print a matrix for home work in a transpose type of way but without pronting the first element of each array (printing place [1] and forward...).
but when i rint it the output shows me that at at the end of every col theres another number thats not suppose to be there (chatgpt says its "garbage") so can someone please help me fix my program in any way cuz i tried for 2 hours and havent figiured it out yet.

-btw one of the things i cant use is indexs in loops and such like int i/j so i have to use only pointers.

#include <stdio.h>

void transpose(int **arr){
  int lenrows = 0;
  int **temp = arr;
  while(*temp != NULL){
    lenrows = lenrows + 1; //ספרתי את שורות הפויינטר ALL
    temp = temp + 1;
  }
  
  int maxCol = 0; //לספירת העמודות
  int **rc = arr;//fst pointer
  while (*rc != NULL) {
    int *colpointer = *rc;   // מצביע לתחילת השורה
    int cur = *colpointer;  // הערך הראשון מציין את האורך של השורה

    if (cur > maxCol){
        maxCol = cur;
    }
    
    rc++;
  }
  
  int col = 0;
  while(col < maxCol){
    int **row = arr;
    while(*row != NULL){
      int *currow = *row;
      int size = *currow;
      
      if (col < size) {
          printf("%-6d", *(currow + col + 1));
        }else{
          printf("      ");
        }
        row = row +1;
    }
    printf("\n");
    col = col +1;
  }
}

int main(){
int A[]={5,-5,14,5,2};
int B[]={3,6,11};
int C[]={4,1,-3,4};
int D[]={6,2,7,1,8,2};
int E[]={2,15};
int F[]={3,4,-2};
int *All[]={A,B,C,D,E,F,NULL};//מערך מצביעים למערכים שבהם יש מספרים

printf("transpose answer: \n");
transpose(All);
}

(i added the og qustion if annyone wanna read)

[strange girder]

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[soft blade]

how is your transpose function supposed to know the lengths of your A, B, . . . F arrays? All is null-terminated, which works in this context, but 0 is a number like any other in the context of an int array

BTW, an "indexless" for loop (depending on how you define "index" IG) could bec int arr[LEN] = {/* whatever */}; for (int* arrPtr = arr; arrPtr < arr + LEN; arrPtr++) { // etc } where at every step of the loop, *arrPtr would be equivalent to arr[i] in a "normal" for loop

[soft blade]

ah, i see, the first array element of each array is the length of the array, and you're supposed to print everything but the length value

in that case, #1386821917998059600 message should be everything you need to know

[queen crest]

?

[acoustic tapir]

What you dont understand?

[queen crest]

Like should i change my code based on what he suggested (the loop without the indexs)?

[acoustic tapir]

The length of each array seems to be the value at index 0 of the individual array. So you could use this to know how many iterations you have to do for each array

[queen crest]

      for(int **ptr = arr; *ptr < arr; ptr++){
      }

[queen crest]

kind of like this would u say ? 👆

[acoustic tapir]

I will look tmr

[soft blade]

ptr < arr + arrLen, where arrLen is the length of the array as an integer, in this case. you want to compare pointers of equal types when you can. in this case, both arr and ptr are of type int**; by dereferencing ptr like that, you're now comparing an int** to an int*. it's probably legal code (in that it will compile), but it won't do what you want it to, because you're comparing an actual pointer with the thing it points to

actually, if this is for the null-terminated array, you'd want to have ptr or ptr != NULL as your condition

[strange girder]

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[queen crest]

!solved