C language pointer space secured [closed]

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Closed 2 years ago.

Two years ago

The following is my program, but the execution results are not working.
I think either the for statement or malloc is strange, but I don't know.

problems:

If you enter one string less than 32 bytes from your keyboard that contains no spaces, create a program that lets malloc fill in the new string by increasing each character to two, display the new string, free up memory, and exit.

Example: "abc" → "aabbcc"

Current State Programs

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

int main(void){

  char*p,q[32];
  inti,k,j;

  printf("Input MOZI:");
  scanf("%s",q);

  i = strlen(q);

  p=(char*)malloc(sizeof(char)*i*2);
  if(p==NULL) {exit(1);}

   printf("%d\n", i);

  for(k=0;k<i;k++){
    for(j=0;j<2;j++){
   *p=*q;
   *p++;
    }
  j = 0;
  ++*q;

  }
  printf("%s\n",p);
  printf("%s\n",q);

  free(p);

}

2022-09-30 18:18

2 Answers

Try changing *p++ and ++*q to p++ and ++q.
You want to proceed with the pointer, not change the contents, right?
* refers to what the pointer points to.
By the way, strictly speaking, ++ should take precedence over *, so there is no problem with *p++, but I think you should remove the asterisk because it is confusing.

2022-09-30 18:18

At a glance, there are the following problems:
I won't point out anything in detail for you to learn, so please look it up yourself.

Not taking action to contain less than 32 bytes of input string to be retrieved
Not taking into account the length of data indicating the end of the string when securing space (@akiraejiri pointed out)
When copying each character, the original pointer itself is rewritten (so subsequent display and space release are not working)
No data is written indicating the destination string termination

2022-09-30 18:18

If you have any answers or tips

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