CODEWITHSUNDEEP
ccharbyte32-bit
user3370603
user3370603

Reputation: 87

Convert char array into int in C

Say you have an char array, which holds 8 bytes. How would you convert that char array into an integer?

I tried using sscanf -

int x;
sscanf(char_array, "%d", &x);

I'm reading bytes from a binary file, storing them into a char array, and then I'm trying to print out an int value based on an offset value.

Upvotes: 0

Views: 7391

Answers (6)

Unda
Unda

Reputation: 1899

If you are reading bytes from a binary file, I suggest to read directly from the file to the integer variable. As followed:

#include        <stdio.h>

int             main() {
  FILE          *file = fopen("myFile", "r");
  int           i;

  if (file) {
    fread(&i, sizeof(i), 1, file);
    printf("%d\n", i);
    fclose(file);
  }
  return (0);
}

But if you cannot get rid of your char array as a source for the conversion, you can use a union to easily perform the conversion:

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

typedef union           u_bytes_to_int_type
{
  int                   value;
  char                  bytes[sizeof(int)];
}                       u_bytes_to_int_type;

void                    reverseArray(char *array, unsigned int const size) {
  char                  tmp;
  unsigned int          reverseIdx;

  for (unsigned int i = 0; i < size / 2; ++i) {
    reverseIdx = size - 1 - i;

    tmp = array[i];
    array[i] = array[reverseIdx];
    array[reverseIdx] = tmp;
  }
}

int                     main() {
  char                  endiannessIsDifferent = 0; // It's up to you how you determine the value of this variable.
  char                  array[sizeof(int)] = {0x2a, 0, 0, 0};
  u_bytes_to_int_type   v;

  memcpy(v.bytes, array, sizeof(array));
  /*
  ** Will reverse the order of the bytes in case the endianness of
  ** the source array is different from the one you need.
  */
  if (endiannessIsDifferent) {
    reverseArray(v.bytes, sizeof(v.bytes));
  }

  printf("%i\n", v.value);
  return (0);
}

Upvotes: 0

Alex
Alex

Reputation: 4473

Is it really char array and not unicode where 2 bytes represent one character? Are you sure result will fit into int? If so, then here is simple solution:

int x=0;
for (int i=0; i<8; i++) x=x*10 + char_array[i]-'0';

Upvotes: 0

Barmak Shemirani
Barmak Shemirani

Reputation: 31599

I think you just want to read hexadecimal numbers which are usually 8 bytes in text form. You can use this:

sscanf(char_array, "%x", &x);

If it's really binary, it would be 4 bytes each and it depends if the machine is little-endian or big-endian. Most computers are little-endian. To make a portable version, you can use these functions:

#ifdef BIG_ENDIAN
#define memcpy_set(buf,v) memcpy(buf, &v, 4)
#else
#define memcpy_set(buf,v) { for (int i = 0; i < 4; i++) buf[i] = v >> (24 - 8 * i); }
#endif
#define memcpy_get(buf) (buf[0] << 24 | buf[1] << 16 | buf[2] << 8 | buf[3])

int main()
{
    if (sizeof(int) != 4)
        return 0;

    char buf[5];
    memset(buf, 0, 5);

    memcpy_set(buf, 0x41424344);
    printf("%s\n", buf);// buf = "ABCD"
    printf("%x\n", memcpy_get(buf)); //0x41424344

    return 0;
}

Upvotes: 1

Liyang Chen
Liyang Chen

Reputation: 122

Combining 8 char into an integer would result in a 64-bit integer, so you may need to declare it as unsigned long long int...

#include <stdio.h>

void showBinary(unsigned long long x) {
    int i;
    for(i = 63; i >= 0; i--) 
        printf("%d", x & (1ULL << i) ? 1 : 0);
    printf("\n");
}

int main(void) {
    char c[8] = {1, 2, 3, 4, 'a', 'b', 'c', 'd'};
    unsigned long long int x = 0;

    int char_i;
    for (char_i = 0; char_i < 8; char_i++) {
        x = (x << 8) | c[char_i];
        showBinary(x);
    }
    printf("Result: x = %lld\n", x);

    return 0;
}

ps. When you read bytes from file, you may need to be careful about the big-endian or little-endian representation.

Upvotes: 1

deluged.oblivion
deluged.oblivion

Reputation: 134

The following converts a 4 byte array (4 chars) into a 32-bit unsigned integer. You should be able to easily extend this to 8 chars (i.e., 64-bit unsigned int).

Iterate array backwards (can do forwards as well) and shift the int representation of the respective character accordingly and fit it into the resultant value.

#include <iostream>
#include <cstdint>

using namespace std;

int main() {
    char arr[] = {0x00, 0x00, 0x1B, 0x1B}; // just for my testing convenience
    uint32_t val = 0;
    for (int i = 3; i >= 0; i--) {
        uint32_t tmp = arr[i];
        int j = 4 - i;
        while (--j) {
            tmp <<= 8;
        }
        val |= tmp;
    }

    cout << val << endl;
}

Upvotes: 1

hpca01
hpca01

Reputation: 378

I generally thought you have to store as a char first, then you cast to int during the println().

Upvotes: -1

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