所以我有一些麻煩得到一個int的字段，然後簽名擴展它。我有一個方法獲取int的字段。符號在C中擴展int int
getField(int value, int hi, int lo);
值是我從一個字段中取得的整數，hi和lo是字段的大小。
所以我可以調用getFieldSignExtended（int值，int hi，int lo）裏面的getField方法，但是如何去擴展它的符號呢？
例如value = 7, hi = 1, lo = 0
getField(7, 1, 0);返回3，因爲以二進制7是111和HI和LO取0領域1.
隨着從getfield命令返回3我得到的值等於爲0x0003。
我到目前爲止的工作都是積極的，但是對消極方面卻大打折扣。當我說「搞砸」時，我的意思是它根本不起作用。所以如果我嘗試在-1上使用它，它會顯示爲一個大的int值而不是負值。
感謝您的幫助！ ：]
編輯：對不起，我混淆了我自己和你們中的一些人的矛盾聲明：P。固定。
您可以採取多種方法。如果你剛剛開始，你可以算術計算現場的位數。所以，舉例來說：
if (value %2 >= 1)
{
// you know that the value has a `1` as the lest significant digit.
// if that's one of the digits you're looking for, you can do something like count++ here
}
else
{
// least significant digit is a '0'
}
然後
if (value % 4 >=2)
{
// you know that the second least significant digit is `1`
// etc.
}
如果你那樣做，你可能會想工作的成某種形式的循環。現在
，更好的方式來做到這一點是使用按位安定，像這樣：
if (value & 8 != 0)
// here you know that the fourth least significant digit (the one representing 8) is 1.
// do a Google search on bitwise anding to get more information.
歡迎來到Stack Overflow。您的回答並不能提供解決問題的完整解決方案，這意味着它的價值有限。我不完全清楚你打算如何擴展你所說的解決問題的方法。一般來說，關於SO的一個好的答案要麼是完整的要麼是容易完成的。如果你對這個問題有一個很好的替代解釋（這確實需要解釋;這不是一個很好的問題），那麼看到你的替代答案會很棒。 –
有很多可怕的行會在這裏閱讀之間。但是，如果getField(7, 1, 0)返回3，並且您需要getFieldSignExtended(15, 2, 0)返回-3和getFieldSignExtended(3, 2, 0)以返回+3，那麼這可能就是您要的。
這個概念是，你把一個n比特字段從比特hi：lo的原始值當作2的補碼來處理。如果n位的第一位是1，那麼你希望n位字段被視爲負數。如果3位字段的第一位是0，那麼你希望它被視爲正數。
#include <assert.h>
#include <limits.h>
#include <stdio.h>
extern int getFieldSignExtended(int value, int hi, int lo);
enum { INT_BITS = CHAR_BIT * sizeof(int) };
int getFieldSignExtended(int value, int hi, int lo)
{
assert(lo >= 0);
assert(hi > lo);
assert(hi < INT_BITS - 1);
int bits = (value >> lo) & ((1 << (hi - lo + 1)) - 1);
if (bits & (1 << (hi - lo)))
return(bits | (~0U << (hi - lo)));
else
return(bits);
}
3個斷言很直接;唯一有爭議的是代碼拒絕處理位31.如果你用hi = 31和lo = 0調用它，那麼移位（hi-lo + 1）太大，行爲不確定。您還會遇到右移一個負數的實現定義的行爲。通過採用無符號整數參數可以解決這些問題，並且如果hi - lo + 1 == INT_BITS不執行&操作。解決這些問題留給讀者來練習。
對bits的賦值將值右移，並用正確的位數屏蔽它。 (1 << (hi - lo + 1)) - 1移位1比字段中的位數多1，然後減1，爲字段中的每個位位置生成一串二進制1。例如，對於hi = 2，lo = 0，這將左移3個位置，產生二進制1000;減1會得到0111，所以選擇正確的3位。因此，bits包含適當的n位整數位。
if測試檢查是否設置了n位整數的最高有效位。如果符號位未設置，我們只需返回值bits。如果符號位被設置，那麼我們有一個棘手的計算來執行 - 這個答案的初稿中（非常）錯誤的。假設我們有一個3位= 101的字段。作爲一個3位2的補碼數，表示-3。我們需要將其擴展到全部爲1的左側，以生成全尺寸的-1。 ~0的值都是位1;當它被左移hi - lo位時，它會爲值的非符號位留下一系列的零。如果向左移動hi - lo + 1，它也可以工作，但是+ 1需要額外的計算，這是不必要的。
我用這個測試工具來確保代碼正常工作。系統測試輸出是嚴格的（小數字）。它確保計算出的值與期望值相匹配。 '窮舉'測試並不完全詳盡;它只測試一個值，而且更適用於觀察問題（例如使用hi = 31和lo = 0在我的機器上給出0的錯誤答案）和模式。
static const struct
{
int value;
int hi;
int lo;
int wanted;
} tests[] =
{
{ 0x0F, 1, 0, -1 },
{ 0x0F, 2, 0, -1 },
{ 0x0F, 2, 1, -1 },
{ 0x0F, 3, 1, -1 },
{ 0x0F, 4, 2, +3 },
{ 0x0F, 5, 0, +15 },
{ 0x0F, 5, 1, +7 },
{ 0x0F, 5, 2, +3 },
{ 0x0F, 5, 3, +1 },
{ 0x0F, 5, 4, 0 },
{ 0x03, 2, 0, +3 },
{ 0xF3, 2, 0, +3 },
{ 0xF3, 3, 0, +3 },
{ 0xF3, 4, 0, -13 },
{ 0xF3, 5, 0, -13 },
{ 0xF3, 6, 0, -13 },
{ 0xF3, 7, 0, -13 },
{ 0xF3, 7, 1, -7 },
{ 0xF3, 7, 2, -4 },
{ 0xF3, 7, 3, -2 },
{ 0xF3, 7, 4, -1 },
{ 0xF3, 8, 0, 0xF3 },
};
enum { NUM_TESTS = sizeof(tests)/sizeof(tests[0]) };
static const char s_pass[] = "== PASS ==";
static const char s_fail[] = "!! FAIL !!";
static void systematic_test(void)
{
int fail = 0;
for (int i = 0; i < NUM_TESTS; i++)
{
char const *pf = s_fail;
int actual = getFieldSignExtended(tests[i].value, tests[i].hi, tests[i].lo);
if (actual == tests[i].wanted)
pf = s_pass;
else
fail++;
printf("%s GFSX(%+4d = 0x%.4X, %d, %d) = %+4d = 0x%.8X (wanted %+4d = 0x%.8X)\n",
pf, tests[i].value, tests[i].value, tests[i].hi, tests[i].lo, actual, actual,
tests[i].wanted, tests[i].wanted);
}
printf("%s\n", (fail == 0) ? s_pass : s_fail);
}
static void exhaustive_test(void)
{
int value = 0x5FA03CE7;
for (int i = 1; i < INT_BITS - 1; i++)
{
for (int j = 0; j < i; j++)
{
int actual = getFieldSignExtended(value, i, j);
printf("%11sGFSX(%d = 0x%X, %2d, %2d) = %+10d = 0x%.8X\n", "",
value, value, i, j, actual, actual);
}
}
}
int main(void)
{
int result1 = getFieldSignExtended(15, 2, 0);
int result2 = getFieldSignExtended(3, 2, 0);
printf("GFSX(15, 2, 0) = %+d = 0x%.8X\n", result1, result1);
printf("GFSX(3, 2, 0) = %+d = 0x%.8X\n", result2, result2);
printf("\nSystematic test\n");
systematic_test();
printf("\nExhaustive test\n");
exhaustive_test();
return(0);
}
這是測試碼的輸出上的窮舉測試之前，加上從窮舉測試輸出的一小部分：
GFSX(15, 2, 0) = -1 = 0xFFFFFFFF
GFSX(3, 2, 0) = +3 = 0x00000003
Systematic test
== PASS == GFSX(+15 = 0x000F, 1, 0) = -1 = 0xFFFFFFFF (wanted -1 = 0xFFFFFFFF)
== PASS == GFSX(+15 = 0x000F, 2, 0) = -1 = 0xFFFFFFFF (wanted -1 = 0xFFFFFFFF)
== PASS == GFSX(+15 = 0x000F, 2, 1) = -1 = 0xFFFFFFFF (wanted -1 = 0xFFFFFFFF)
== PASS == GFSX(+15 = 0x000F, 3, 1) = -1 = 0xFFFFFFFF (wanted -1 = 0xFFFFFFFF)
== PASS == GFSX(+15 = 0x000F, 4, 2) = +3 = 0x00000003 (wanted +3 = 0x00000003)
== PASS == GFSX(+15 = 0x000F, 5, 0) = +15 = 0x0000000F (wanted +15 = 0x0000000F)
== PASS == GFSX(+15 = 0x000F, 5, 1) = +7 = 0x00000007 (wanted +7 = 0x00000007)
== PASS == GFSX(+15 = 0x000F, 5, 2) = +3 = 0x00000003 (wanted +3 = 0x00000003)
== PASS == GFSX(+15 = 0x000F, 5, 3) = +1 = 0x00000001 (wanted +1 = 0x00000001)
== PASS == GFSX(+15 = 0x000F, 5, 4) = +0 = 0x00000000 (wanted +0 = 0x00000000)
== PASS == GFSX( +3 = 0x0003, 2, 0) = +3 = 0x00000003 (wanted +3 = 0x00000003)
== PASS == GFSX(+243 = 0x00F3, 2, 0) = +3 = 0x00000003 (wanted +3 = 0x00000003)
== PASS == GFSX(+243 = 0x00F3, 3, 0) = +3 = 0x00000003 (wanted +3 = 0x00000003)
== PASS == GFSX(+243 = 0x00F3, 4, 0) = -13 = 0xFFFFFFF3 (wanted -13 = 0xFFFFFFF3)
== PASS == GFSX(+243 = 0x00F3, 5, 0) = -13 = 0xFFFFFFF3 (wanted -13 = 0xFFFFFFF3)
== PASS == GFSX(+243 = 0x00F3, 6, 0) = -13 = 0xFFFFFFF3 (wanted -13 = 0xFFFFFFF3)
== PASS == GFSX(+243 = 0x00F3, 7, 0) = -13 = 0xFFFFFFF3 (wanted -13 = 0xFFFFFFF3)
== PASS == GFSX(+243 = 0x00F3, 7, 1) = -7 = 0xFFFFFFF9 (wanted -7 = 0xFFFFFFF9)
== PASS == GFSX(+243 = 0x00F3, 7, 2) = -4 = 0xFFFFFFFC (wanted -4 = 0xFFFFFFFC)
== PASS == GFSX(+243 = 0x00F3, 7, 3) = -2 = 0xFFFFFFFE (wanted -2 = 0xFFFFFFFE)
== PASS == GFSX(+243 = 0x00F3, 7, 4) = -1 = 0xFFFFFFFF (wanted -1 = 0xFFFFFFFF)
== PASS == GFSX(+243 = 0x00F3, 8, 0) = +243 = 0x000000F3 (wanted +243 = 0x000000F3)
== PASS ==
Exhaustive test
GFSX(1604336871 = 0x5FA03CE7, 1, 0) = -1 = 0xFFFFFFFF
GFSX(1604336871 = 0x5FA03CE7, 2, 0) = -1 = 0xFFFFFFFF
GFSX(1604336871 = 0x5FA03CE7, 2, 1) = -1 = 0xFFFFFFFF
GFSX(1604336871 = 0x5FA03CE7, 3, 0) = +7 = 0x00000007
GFSX(1604336871 = 0x5FA03CE7, 3, 1) = +3 = 0x00000003
GFSX(1604336871 = 0x5FA03CE7, 3, 2) = +1 = 0x00000001
GFSX(1604336871 = 0x5FA03CE7, 4, 0) = +7 = 0x00000007
GFSX(1604336871 = 0x5FA03CE7, 4, 1) = +3 = 0x00000003
GFSX(1604336871 = 0x5FA03CE7, 4, 2) = +1 = 0x00000001
GFSX(1604336871 = 0x5FA03CE7, 4, 3) = +0 = 0x00000000
GFSX(1604336871 = 0x5FA03CE7, 5, 0) = -25 = 0xFFFFFFE7
GFSX(1604336871 = 0x5FA03CE7, 5, 1) = -13 = 0xFFFFFFF3
GFSX(1604336871 = 0x5FA03CE7, 5, 2) = -7 = 0xFFFFFFF9
GFSX(1604336871 = 0x5FA03CE7, 5, 3) = -4 = 0xFFFFFFFC
GFSX(1604336871 = 0x5FA03CE7, 5, 4) = -2 = 0xFFFFFFFE
GFSX(1604336871 = 0x5FA03CE7, 6, 0) = -25 = 0xFFFFFFE7
GFSX(1604336871 = 0x5FA03CE7, 6, 1) = -13 = 0xFFFFFFF3
GFSX(1604336871 = 0x5FA03CE7, 6, 2) = -7 = 0xFFFFFFF9
GFSX(1604336871 = 0x5FA03CE7, 6, 3) = -4 = 0xFFFFFFFC
GFSX(1604336871 = 0x5FA03CE7, 6, 4) = -2 = 0xFFFFFFFE
GFSX(1604336871 = 0x5FA03CE7, 6, 5) = -1 = 0xFFFFFFFF
...
GFSX(1604336871 = 0x5FA03CE7, 29, 28) = +1 = 0x00000001
GFSX(1604336871 = 0x5FA03CE7, 30, 0) = -543146777 = 0xDFA03CE7
GFSX(1604336871 = 0x5FA03CE7, 30, 1) = -271573389 = 0xEFD01E73
GFSX(1604336871 = 0x5FA03CE7, 30, 2) = -135786695 = 0xF7E80F39
GFSX(1604336871 = 0x5FA03CE7, 30, 3) = -67893348 = 0xFBF4079C
GFSX(1604336871 = 0x5FA03CE7, 30, 4) = -33946674 = 0xFDFA03CE
GFSX(1604336871 = 0x5FA03CE7, 30, 5) = -16973337 = 0xFEFD01E7
GFSX(1604336871 = 0x5FA03CE7, 30, 6) = -8486669 = 0xFF7E80F3
GFSX(1604336871 = 0x5FA03CE7, 30, 7) = -4243335 = 0xFFBF4079
GFSX(1604336871 = 0x5FA03CE7, 30, 8) = -2121668 = 0xFFDFA03C
GFSX(1604336871 = 0x5FA03CE7, 30, 9) = -1060834 = 0xFFEFD01E
GFSX(1604336871 = 0x5FA03CE7, 30, 10) = -530417 = 0xFFF7E80F
GFSX(1604336871 = 0x5FA03CE7, 30, 11) = -265209 = 0xFFFBF407
GFSX(1604336871 = 0x5FA03CE7, 30, 12) = -132605 = 0xFFFDFA03
GFSX(1604336871 = 0x5FA03CE7, 30, 13) = -66303 = 0xFFFEFD01
GFSX(1604336871 = 0x5FA03CE7, 30, 14) = -33152 = 0xFFFF7E80
GFSX(1604336871 = 0x5FA03CE7, 30, 15) = -16576 = 0xFFFFBF40
GFSX(1604336871 = 0x5FA03CE7, 30, 16) = -8288 = 0xFFFFDFA0
GFSX(1604336871 = 0x5FA03CE7, 30, 17) = -4144 = 0xFFFFEFD0
GFSX(1604336871 = 0x5FA03CE7, 30, 18) = -2072 = 0xFFFFF7E8
GFSX(1604336871 = 0x5FA03CE7, 30, 19) = -1036 = 0xFFFFFBF4
GFSX(1604336871 = 0x5FA03CE7, 30, 20) = -518 = 0xFFFFFDFA
GFSX(1604336871 = 0x5FA03CE7, 30, 21) = -259 = 0xFFFFFEFD
GFSX(1604336871 = 0x5FA03CE7, 30, 22) = -130 = 0xFFFFFF7E
GFSX(1604336871 = 0x5FA03CE7, 30, 23) = -65 = 0xFFFFFFBF
GFSX(1604336871 = 0x5FA03CE7, 30, 24) = -33 = 0xFFFFFFDF
GFSX(1604336871 = 0x5FA03CE7, 30, 25) = -17 = 0xFFFFFFEF
GFSX(1604336871 = 0x5FA03CE7, 30, 26) = -9 = 0xFFFFFFF7
GFSX(1604336871 = 0x5FA03CE7, 30, 27) = -5 = 0xFFFFFFFB
GFSX(1604336871 = 0x5FA03CE7, 30, 28) = -3 = 0xFFFFFFFD
GFSX(1604336871 = 0x5FA03CE7, 30, 29) = -2 = 0xFFFFFFFE
「符號擴展用零」是自相矛盾的。 '符號擴展'表示'將符號位（0或1）複製到擴展名';只用零來擴展它，如果類型是無符號和擴展的話會發生什麼。 –
您的問題未指定，因此有可能被標記爲「關閉」爲「不是真正的問題」。另外，你嘗試過什麼方法？ – Kaz
你的例子中的符號擴展可能意味着返回-1，因爲如果我們從'... 0000111'中取出'11'位，我們必須將'11'解釋爲一個兩位二進制補碼值，它代表-1 。在C中，在一個二進制補碼平臺上，一個帶符號的結構/聯合位域是兩位寬，其中保存「11」的結果爲-1。 – Kaz