[英]c; converting 2 bytes to short and vice versa
我想將2到2的字節數組bytes1
(小端)轉換成短整數數組,反之亦然。 我期望得到最終數組bytes2
,它等於初始數組bytes1
。 我有這樣的代碼:
int i = 0;
int j = 0;
char *bytes1;
char *bytes2;
short *short_ints;
bytes1 = (char *) malloc( 2048 );
bytes2 = (char *) malloc( 2048 );
short_ints = (short *) malloc( 2048 );
for ( i=0; i<2048; i+=2)
{
short_ints[j] = bytes1[i+1] << 8 | bytes1[i] ;
j++;
}
j = 0;
for ( i=0; i<2048; i+=2)
{
bytes2[i+1] = (short_ints[j] >> 8) & 0xff;
bytes2[i] = (short_ints[j]) ;
j++;
}
j = 0;
現在,誰能告訴我,為什么我沒有得到bytes2
陣列,完全一樣bytes1
? 以及如何正確執行此操作?
建議2個功能。 將所有合並和提取都以無符號方式進行,以消除符號位short
(可能是char
。
符號位是OP的代碼最大的問題。 short_ints[j] = bytes1[i+1] << 8 | bytes1[i] ;
如果將bytes1[i]
轉換為int
則符號可能會擴展。
另外(short_ints[j] >> 8)
擴展符號。
// Combine every 2 char (little endian) into 1 short
void charpair_short(short *dest, const char *src, size_t n) {
const unsigned char *usrc = (const unsigned char *) src;
unsigned short *udest = (unsigned short *) dest;
if (n % 2) Handle_OddError();
n /= 2;
while (n-- > 0) {
*udest = *usrc++;
*udest += *usrc++ * 256u;
udest++;
}
}
// Break every short into 2 char (little endian)
void short_charpair(char *dest, const short *src, size_t n) {
const unsigned short *usrc = (const unsigned short *) src;
unsigned char *udest = (unsigned char *) dest;
if (n % 2) Handle_OddError();
n /= 2;
while (n-- > 0) {
*udest++ = (unsigned char) (*usrc);
*udest++ = (unsigned char) (*usrc / 256u);
usrc++;
}
}
int main(void) {
size_t n = 2048; // size_t rather than int has advantages for array index
// Suggest code style: type *var = malloc(sizeof(*var) * N);
// No casting of return
// Use sizeof() with target pointer name rather than target type.
char *bytes1 = malloc(sizeof * bytes1 * n);
Initialize(bytes, n); //TBD code for OP-best to not work w/uninitialized data
// short_ints = (short *) malloc( 2048 );
// This is weak as `sizeof(short)!=2` is possible
short *short_ints = malloc(sizeof * short_ints * n/2);
charpair_short(short_ints, bytes1, n);
char *bytes2 = malloc(sizeof * bytes2 * n);
short_charpair(bytes2, short_ints, n);
compare(bytes1, bytes2, n); // TBD code for OP
// epilogue
free(bytes1);
free(short_ints);
free(bytes2);
return 0;
}
避免使用union
方法,因為這取決於平台字節序。
這是一個程序,演示您正在遇到與帶符號移位整數值相關的問題。
#include <stdio.h>
#include <stdlib.h>
void testCore(char bytes1[],
char bytes2[],
short short_ints[],
int size)
{
int i = 0;
int j = 0;
for ( i=0; i<size; i+=2)
{
short_ints[j] = bytes1[i+1] << 8 | bytes1[i] ;
j++;
}
j = 0;
for ( i=0; i<size; i+=2)
{
bytes2[i+1] = (short_ints[j] >> 8) & 0xff;
bytes2[i] = (short_ints[j]) ;
j++;
}
for ( i=0; i<size; ++i)
{
if ( bytes1[i] != bytes2[i] )
{
printf("%d-th element is not equal\n", i);
}
}
}
void test1()
{
char bytes1[4] = {-10, 0, 0, 0};
char bytes2[4];
short short_ints[2];
testCore(bytes1, bytes2, short_ints, 4);
}
void test2()
{
char bytes1[4] = {10, 0, 0, 0};
char bytes2[4];
short short_ints[2];
testCore(bytes1, bytes2, short_ints, 4);
}
int main()
{
printf("Calling test1 ...\n");
test1();
printf("Done\n");
printf("Calling test2 ...\n");
test2();
printf("Done\n");
return 0;
}
程序輸出:
Calling test1 ... 1-th element is not equal Done Calling test2 ... Done
宇大
這是一個適用於我的testCore
版本:
void testCore(char bytes1[],
char bytes2[],
short short_ints[],
int size)
{
int i = 0;
int j = 0;
unsigned char c1;
unsigned char c2;
unsigned short s;
for ( i=0; i<size; i+=2)
{
c1 = bytes1[i];
c2 = bytes1[i+1];
short_ints[j] = (c2 << 8) | c1;
j++;
}
j = 0;
for ( i=0; i<size; i+=2)
{
s = short_ints[j];
s = s >> 8;
bytes2[i+1] = s;
bytes2[i] = short_ints[j] & 0xff;
j++;
}
for ( i=0; i<size; ++i)
{
if ( bytes1[i] != bytes2[i] )
{
printf("%d-th element is not equal\n", i);
}
}
}
經過測試:
char bytes1[4] = {-10, 0, 25, -4};
和
char bytes1[4] = {10, -2, 25, 4};
好吧,您需要的是UNION:
#include <stdio.h>
#include <string.h>
union MyShort {
short short_value;
struct {
char byte1;
char byte2;
};
};
int main(int argc, const char * argv[])
{
char a[4]="abcd";
char b[4]="1234";
short c[5]; c[4]=0;
union MyShort d;
for (int i = 0; i<4; i++) {
d.byte1 = a[i];
d.byte2 = b[i];
c[i] = d.short_value;
}//next i
printf("%s\n", (char*)c);
return 0;
}
結果應為a1b2c3d4。
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