如何使用int指针访问char数组？

Question

Hi how to access character array using integer point.

char arr[10] = {'1','2','3','4','5','6','7','8','9','10'};
int *ptr;

How i can print values of 'arr' using pointer ptr?

Answer 1

It is a little unclear what your goal is, but trying to print out a character array with an integer pointer is a bit like trying to get to the second step taking four-steps at a time. When you tell the compiler that you would like to reference a memory address with an integer pointer, the compiler knows that an integer is sizeof (int) bytes (generally 4-bytes on x86/x86_64). So attempting to access each element in a character array with an integer pointer and normal pointer arithmetic wouldn't work. (you would be advancing 4-bytes at a time).

However printing the character array though an integer pointer is possible if you use the integer pointer for the starting address of the array and advance the pointer by the number of characters in the array by casting back to char . While it is doubtful this is your goal, the plain statement of your question seems to suggest it. To accomplish this, you could:

#include <stdio.h>

int main (void)
{
    char arr[] = {'1','2','3','4','5','6','7','8','9'};
    int *ptr = (int *)arr;
    unsigned int i;

    for (i = 0; i < sizeof arr; i++)
        printf (" %c", (*(char *)ptr + i));
    putchar ('\n');

    return 0;
}

Output

$ ./bin/char_array_int_ptr
 1 2 3 4 5 6 7 8 9

Note: your original initialization of your array with a character '10' was invalid. If this was an assignment, it is likely intended to expose you to how pointer arithmetic is influenced by type and the ability to cast from and to type char (without violating strict aliasing rules)

Answer 2

If you are just after the integer values you can print out the characters as integers

for (i = 0; i < sizeof(arr)/sizeof(arr[0]); ++i)
{
  printf( "%d ", arr[i] );
}

Answer 3

Using a pointer of the wrong data type to access anything is undefined behavior, thus making it not something you want to do. If you want to cast a char to an integer, you can do that. If you want to print the integer value of a char, you can do that too.

But using a pointer type integer to access a char array is undefined behavior.

Answer 4

Simply assign the array name to the pointer (using some casting):

int* ptr = (int*)arr;

An array variable is actually just a pointer to the array's first element, so this code is equivalent to:

int* ptr = (int*)&arr[0];

As was mentioned in this comments, you can get yourself into some nasty trouble doing this. On most modern 32-bit systems, int is 4 times as wide as char , so incrementing ptr will jump forward 4 elements in arr , and you'll get 4 chars interpreted as a single int when you dereference ptr . If you're not extremely careful, this will cause segfaults and/or undefined behavior. In fact, you'll probably get undefined behavior even if you are careful.

That said, this is almost certainly not you want to do. It sounds like you want to interpret the values in arr as integers. To do that, you first have to understand that char s in C and C++ are really not any different from int s (except for their memory size and the range of values they can represent). '0' is equivalent to 48 (the ASCII ordinal for the character 0 ). So, to interpret a single digit as an integer, do something like this:

int x = arr[i]-'0';

Where i is whatever index in the array you want to access. If you want to interpret multiple-digit strings as integers, atoi is the tool for the job:

#include <stdlib.h>
#include <stdio.h>
int main() {
  char* s = "12345";
  int x = atoi(s);
  printf("%d\n", x); // 12345
  return 0;
}