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我有一小段代碼嘗試演示各種標準整數 類型，並顯示內存似乎首先被分配到堆棧中，並且最大的數據類型爲 。我寫的東西似乎是合理的，直到最後一行，我得到的輸出字節太多了。我真的只想（並期望）只能得到一個字節。一次從一個字節獲取變量存儲器地址的數據
這樣：
#include <stdint.h> /* defines the standard integer types */
#include <stdio.h> /* defines all the IO functions */
#include <stddef.h> /* standard definitions */
#include <stdlib.h> /* standard library functions */
int main (int argc, char *argv[]) {
int8_t little_i = 11;
int16_t med_i = 12062;
int32_t norm_i = 1991;
int64_t big_i = -3245321806;
/* these next items are pointers */
int8_t *little_p;
printf ("here is a little 8-bit int : little_i = %02x\n", little_i);
printf ("here is a medium 16-bit int : med_i = %04x\n", med_i);
printf ("here is a normal 32-bit int : norm_i = %08x\n", norm_i);
printf ("here is a big 64-bit int : big_i = %016lx\n", big_i);
printf ("addr of little_i is %p" , &little_i);
printf (" and the size is %d\n" , sizeof(little_i));
printf ("addr of med_i is %p" , &med_i);
printf (" and the size is %d\n" , sizeof(med_i));
printf ("addr of norm_i is %p" , &norm_i);
printf (" and the size is %d\n" , sizeof(norm_i));
printf ("addr of big_i is %p" , &big_i);
printf (" and the size is %d\n" , sizeof(big_i));
/* tell us about the pointer */
printf ("\n--------------------------------------------------\n");
printf (" size of the pointer little_p is %d\n" , sizeof(little_p));
printf (" the address of the pointer itself is %p\n", &little_p);
little_p = &little_i;
printf ("\nthe pointer little_p now contains the addr %p\n", little_p);
printf ("The data there, in memory, at the addr %p is 0x%02xh\n",
little_p, *little_p);
/* we can point a pointer anywhere we want */
little_p = (void *) &big_i;
printf ("\n\nthe pointer little_p now has the addr %p\n", little_p);
printf ("The data there, in memory, at the addr is 0x%02xh\n",
*((int8_t*) little_p));
return (EXIT_SUCCESS);
}
我得到的輸出看起來不錯，直到最後一行：
$ ./dtypes
here is a little 8-bit int : little_i = 0b
here is a medium 16-bit int : med_i = 2f1e
here is a normal 32-bit int : norm_i = 000007c7
here is a big 64-bit int : big_i = ffffffff3e9051b2
addr of little_i is ffffffff7ffff56b and the size is 1
addr of med_i is ffffffff7ffff568 and the size is 2
addr of norm_i is ffffffff7ffff564 and the size is 4
addr of big_i is ffffffff7ffff558 and the size is 8
--------------------------------------------------
size of the pointer little_p is 8
the address of the pointer itself is ffffffff7ffff550
the pointer little_p now contains the addr ffffffff7ffff56b
The data there, in memory, at the addr ffffffff7ffff56b is 0x0bh
the pointer little_p now has the addr ffffffff7ffff558
The data there, in memory, at the addr is 0xffffffffh
我期望看到的最後一行只是一個字節0xffh卻得到了一個完整的32位整數大小的結果。我認爲將little_p強制轉換爲int8_t *類型的指針會起到訣竅的作用。
我錯過了一些明顯的東西嗎？
-----更新就在我張貼了這個問題：
我改變的最後一行有一個投給uint8_t *：
printf ("The data there, in memory, at the addr is 0x%02xh\n",
*((uint8_t*) little_p));
這使我的輸出：
The data there, in memory, at the addr is 0xffh
主要是因爲內存中的字節一次無符號8位。
似乎合理。
-----------進一步編輯提取和從64位整數打印字節------
下面是一個更新的代碼塊：
$ cat dtypes.c
#include <stdint.h> /* defines the standard integer types */
#include <stdio.h> /* defines all the IO functions */
#include <stddef.h> /* standard definitions */
#include <stdlib.h> /* standard library functions */
int main (int argc, char *argv[]) {
int8_t little_i = 11;
int16_t med_i = 12062;
int32_t norm_i = 1991;
int64_t big_i = -3245321806;
void *little_p; /* void datatype is no datatype at all really. */
printf ("here is a little 8-bit int : little_i = %02x\n", little_i);
printf ("here is a medium 16-bit int : med_i = %04x\n", med_i);
printf ("here is a normal 32-bit int : norm_i = %08x\n", norm_i);
printf ("here is a big 64-bit int : big_i = %016lx\n", big_i);
printf ("addr of little_i is %p" , &little_i);
printf (" and the size is %d\n" , sizeof(little_i));
printf ("addr of med_i is %p" , &med_i);
printf (" and the size is %d\n" , sizeof(med_i));
printf ("addr of norm_i is %p" , &norm_i);
printf (" and the size is %d\n" , sizeof(norm_i));
printf ("addr of big_i is %p" , &big_i);
printf (" and the size is %d\n" , sizeof(big_i));
/* tell us about the pointer */
printf ("\n--------------------------------------------------\n");
printf (" size of the pointer little_p is %d\n" , sizeof(little_p));
printf (" the address of the pointer itself is %p\n", &little_p);
/* we can point a pointer anywhere we want */
little_p = (void *) &big_i;
printf ("\n\nthe pointer little_p now has the addr %p\n", little_p);
printf ("The data there, in memory, at the addr is 0x%02xh\n",
*((uint8_t*) little_p));
printf (" at addr+1 is 0x%02xh\n",
*((uint8_t*) little_p+1));
printf (" at addr+2 is 0x%02xh\n",
*((uint8_t*) little_p+2));
printf (" at addr+3 is 0x%02xh\n",
*((uint8_t*) little_p+3));
printf (" at addr+4 is 0x%02xh\n",
*((uint8_t*) little_p+4));
printf (" at addr+5 is 0x%02xh\n",
*((uint8_t*) little_p+5));
printf (" at addr+6 is 0x%02xh\n",
*((uint8_t*) little_p+6));
printf (" at addr+7 is 0x%02xh\n",
*((uint8_t*) little_p+7));
return (EXIT_SUCCESS);
}
這給了非常好的結果：
$ ./dtypes
here is a little 8-bit int : little_i = 0b
here is a medium 16-bit int : med_i = 2f1e
here is a normal 32-bit int : norm_i = 000007c7
here is a big 64-bit int : big_i = ffffffff3e9051b2
addr of little_i is ffffffff7ffff56b and the size is 1
addr of med_i is ffffffff7ffff568 and the size is 2
addr of norm_i is ffffffff7ffff564 and the size is 4
addr of big_i is ffffffff7ffff558 and the size is 8
--------------------------------------------------
size of the pointer little_p is 8
the address of the pointer itself is ffffffff7ffff550
the pointer little_p now has the addr ffffffff7ffff558
The data there, in memory, at the addr is 0xffh
at addr+1 is 0xffh
at addr+2 is 0xffh
at addr+3 is 0xffh
at addr+4 is 0x3eh
at addr+5 is 0x90h
at addr+6 is 0x51h
at addr+7 is 0xb2h
一切似乎好這裏。
我剛剛發現我自己。必須記得在早上製作第一杯咖啡，然後在我發佈前思考..但這對我來說可能是合理的教訓。內存不是簽名數據。它只是記憶。 –
也... ..愛你的用戶名。是的，我記得當我們有大的電腦控制室和大量的das blinken燈:-) –
@paullanken：還要注意，用十六進制打印字符大小數量的格式是'％hhx'，而不是'％xh'。無論簽名如何，更改格式也會產生預期輸出。 –