[英]How to store odd and even position bits to another pointer using C?
我有兩個指針。 一個是輸入ptr,另一個是輸出。
我的輸入指針包含二進制值
15.................0
0011 0110 1111 0111 - unsigned 16 bit
1011 0100 1011 1100 - unsigned 16 bit
首先,我想從兩行中獲取偶數位,並且必須將其填充到輸出指針中。
o/p 指針將是
0110 0110 0110 1111 (even position bits from both)..from right to left and assume 0 to 15.
1100 1110 0101 1101 (odd position bits from both)
如何通過使用指針來做到這一點? 我對指針很陌生
轉換您的代碼以使用指針通常非常簡單。
舉個例子:
uint16_t compute(uint16_t a, uint16_t b)
{
return ((a | b) & MY_BIT_MASK) >> (a & 2); // some arbitrary operations
}
uint16_t compute_p(uint16_t *a, uint16_t *b)
{
return ((*a | *b) & MY_BIT_MASK) >> (*a & 2); // some arbitrary operations via pointers
}
如果您向我們展示您的實際代碼,我們或許可以為您提供更多幫助。 但基本上,當你有一個指向 uint16_t 的指針時,你可以通過兩種方式使用它:
uint16_t *p;
p <- to mean the pointers value, which is an address, you assign it to other pointers, or increment it, so it would point to the next uint16_t in memory
*p <- to use it in expressions on place of a uint16_t, this way your code will each time fetch the actual 16 bit value pointed to by p
p[i] <- with int i, This is basically a synonym to *(p+i) as long as i is non-negative
*(p + i) <- with int i, This is also the value of 16 bit integer where p points to, or the next, or previous etc 16 bit value. In this form, negative i can be used as well
p - q <- with uint16* q the difference of the pointers, in case of uint16_t * types, this basically means "how many 16 but ints I can store and addresses q, q+1, q+2, q+3 .... p-1", assuming p>q
據我所知,你不能用 C 中的指針做你想做的事。這是因為,據我所知,你不能在 C 中創建指向位的指針,它們只能指向字節。
我能想到的一種解決方案是使用位掩碼。 通過屏蔽單個位,將它們移動到您想要的位置,並將它們分配到一個新的空白變量中,可以實現您想要的。 在這里,檢查以下內容:
#define bitmask(_x_) (1 << (_x_))
#include <stdio.h>
#include <inttypes.h>
int main( ){
uint16_t in1 = 14071; // 0011 0110 1111 0111
uint16_t in2 = 46268; // 1011 0100 1011 1100
uint16_t out1 = 0;
uint16_t out2 = 0;
for ( int i = 0; i < 8; i++ ) {
out1 |= ( in1 & bitmask( i * 2 ) ) >> i;
out1 |= ( ( in2 & bitmask( i * 2 ) ) >> i ) << 8;
out2 |= ( in1 & bitmask( i * 2 + 1 ) ) >> ( i + 1 );
out2 |= ( ( in2 & bitmask( i * 2 + 1 ) ) >> ( i + 1 ) ) << 8;
}
printf( "out1: %u\nout2: %u", out1, out2 );
return 0;
}
輸出是:
out1: 26223
out2: 52829
其中具有您想要的基數 2 的形式。
編輯:具有相同輸出的替代版本,這可能對讀者更友好,我不知道性能方面:
#define bitMASK(maskee,bit) (((maskee) >> (bit)) & 1)
#include <stdio.h>
#include <inttypes.h>
int main( ){
uint16_t in1 = 14071; // 0011 0110 1111 0111
uint16_t in2 = 46268; // 1011 0100 1011 1100
uint16_t out1 = 0;
uint16_t out2 = 0;
for ( int i = 0; i < 8; i++ ) {
out1 |= bitMASK( in1, 2 * i ) << i;
out1 |= bitMASK( in2, 2 * i ) << ( 8 + i );
out2 |= bitMASK( in1, 2 * i + 1 ) << i;
out2 |= bitMASK( in2, 2 * i + 1 ) << ( 8 + i );
}
printf( "out1: %u\nout2: %u", out1, out2 );
return 0;
}
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