我想在C中做一些操作。直到现在,我所做的一切都是正确的,但最终测试表明存在错误。对我来说功能看起来不错,但是......输出与预期不同。位操作出错C
#include <stdint.h>
#include "PETER_interface.h"
const int BITS_IN_BYTE = 8;
static int g_capacity = 0;
static int g_size = 0;
static container_id_t indexOfTable = 0;
uint8_t* storages[4] = {0 ,0 ,0 ,0};
err_t PETER_new(container_id_t* id, int num_of_bits) {
storages[indexOfTable] = (uint8_t*)(malloc((num_of_bits/BITS_IN_BYTE)
+ (num_of_bits % BITS_IN_BYTE) ? 1 : 0));
g_capacity = num_of_bits;
g_size = ((num_of_bits/BITS_IN_BYTE) + (num_of_bits % BITS_IN_BYTE) ?
1 : 0);
*id = indexOfTable;
indexOfTable++;
return E_OK;
}
err_t PETER_resize(container_id_t id, int bit) {
return E_NOT_IMPLEMENTED;
}
err_t PETER_delete(container_id_t id) {
return E_NOT_IMPLEMENTED;
}
err_t PETER_deleteAll() {
return E_NOT_IMPLEMENTED;
}
err_t PETER_set(container_id_t id, int bit) {
uint8_t temp = *(storages[id])/BITS_IN_BYTE;
temp |= (1 << (bit % BITS_IN_BYTE));
*(storages[id]) = temp;
return E_OK;
}
err_t PETER_clear(container_id_t id, int bit) {
uint8_t temp = *(storages[id])/BITS_IN_BYTE;
temp &= ~(1 << (bit % BITS_IN_BYTE));
*(storages[id]) = temp;
return E_OK;
}
err_t PETER_invert(container_id_t id, int bit) {
uint8_t temp = *(storages[id])/BITS_IN_BYTE;
temp ^= (1 << (bit % BITS_IN_BYTE));
*(storages[id]) = temp;
return E_OK;
}
err_t PETER_capacity(container_id_t id, int *capacity) {
*capacity = g_capacity;
return E_OK;
}
err_t PETER_storageSize(container_id_t id, int *size) {
*size = g_size;
return E_OK;
}
err_t PETER_get(container_id_t id, int flag_no, int *return_value) {
*return_value = (*(storages[id]) >> (flag_no % BITS_IN_BYTE)) & 1;
return E_OK;
}
我的测试是这样的:
int ok = 1;
const int size = 8;
container_id_t id;
int val;
OK( new(&id,size), E_OK );
OK( capacity(id, &val), E_OK );
OK( val, size);
OK( storageSize(id, &val), E_OK );
OK( val, 1);
printf("storageSize: %d\n", val);
OK( set(id, 4), E_OK );
OK( get(id, 4, &val), E_OK );
printf("set: %d\n", val);
OK( val, 1);
OK( clear(id, 4), E_OK );
OK( get(id, 4, &val), E_OK );
printf("clear: %d\n", val);
OK( val, 0);
OK( invert(id, 4), E_OK );
OK( get(id, 4, &val), E_OK );
printf("invert: %d\n", val);
OK( val, 1);
OK( invert(id, 4), E_OK );
OK( get(id, 4, &val), E_OK );
printf("invert: %d\n", val);
OK( val, 0);
// not yet implemented
OK( resize(id, 2*size),E_NOT_IMPLEMENTED );
OK( delete(id), E_NOT_IMPLEMENTED );
OK( deleteAll(), E_NOT_IMPLEMENTED );
if(ok == 1) {
OUT_GREEN();
printf("%s: ok", module_name);
OUT_WHITE();
}
输出,我得到如下:
Run tests
storageSize: 1
set: 1
clear: 0
invert: 1
invert: 1
fail line 84 (which is second invert)
我试图找到哪里是错误的,但我什么也看不到。也许你们和女孩们可以看到?
什么是所有'PETER_xxx'的东西?这与测试计划中的调用有什么关系? – Barmar
@Barmar你可以离开它。这就是我的模块名称 – gawron103
在处理位操作时,通常应该使用'unsigned int'。 – Barmar