Reputation: 121
I have an array and I want to convert it to an int32_t array.
I tried the code below.
int32_t const_data[11];
int8_t buffer[44];
int k = 0;
while (k < 11) {
for (int j = 0; j < 44; j++) {
const_data[k] = bf.buffer[j];
k++;
}
}
Upvotes: 3
Views: 2846
Reputation: 1371
The easiest and most straight forward way is to use a union
#define array_size_int32 11
#define array_size_int8 44
typedef union{
int32_t const_data[array_size_int32];
int8_t buffer[array_size_int8];
}my_union_t;
Example usage:
/* initialize union members */
my_union_t my_union = {
.const_data = {
0x12345678,
0x12345678,
0x12345678,
0x12345678,
0x12345678,
0x12345678,
0x12345678,
0x12345678,
0x12345678,
0x12345678,
0x12345678,
},
};
Example way to print:
uint8_t i;
for(i = 0; i < array_size_int8; i++){
/* mask off sign extension bits */
printf("my_union.buffer[%d] = %x\n", i, my_union.buffer[i] & 0xff);
}
You can try the code out here
EDIT:
I should add that this works because the memory size needed to allocate either array is the same and you'll run in to problems if you change the #define
's without taking that in to consideration.
For example,
#define array_size_int32 10 //40 bytes
#define array_size_int8 45 //45 bytes
Upvotes: 4
Reputation: 3277
Below is solution which I used in past for similar problem.
#include <stdio.h>
#include <inttypes.h>
int main(){
int8_t input[] = {0x01,0x02,0x03,0x04,0x05,0x06,0x07,0x08};
int32_t output[10];
int count=0;
for(;count<sizeof(input);count+=sizeof(int32_t)/sizeof(int8_t)){
output[count] = (int32_t)(*(int32_t*)(&input[count]));
printf("%x:%x\n",input[count],output[count]);
}
return 0;
}
Note: As pointed out by @tadman in comment. You also need to consider endianness of your platform. This solution works well for my problem on platform I was working. Based on your platform you may need to tweak this.
Upvotes: -1