庫中的函數源碼

1. eclipse中如何用快捷鍵查找java類庫中的源碼

這個分兩步：

1、滑鼠放到你要看的函數上面，按ctrl鍵，可以看到定義的提示如下圖：

2. sql server 中如何查看自定義函數的源代碼

可按如下方法查詢，以sqlserver2008為例：
1、登錄SQL
Server
Management
Studio。
2、展開左邊的樹，先在資料庫中找到自己創建自定義函數的庫，如資料庫-系統資料庫-master。
3、依次點擊可編程性-函數-標量值函數，如曾經創建過一個叫「fn_myget」的自定義函數，就能看見。
4、右鍵此函數，點擊編輯，就能看到這個函數的源代碼。
5、代碼如圖，紅框部分即為源代碼。

3. 如何看c語言標准庫函數的源代碼

很遺憾,標准庫中的函數結合了系統,硬體等的綜合能力,是比較近機器的功能實現,所以大部分是用匯編完成的,而且已經導入到了lib和dll里了,就是說,他們已經被編譯好了,似乎沒有代碼的存在了.
能看到的也只有dll中有多少函數被共享.
第三方可能都是dll,因為上面也說了,dll是編譯好的,只能看到成品,就可以隱藏代碼,保護自己的知識產權,同時也是病毒的歸宿...... 當然,除了DLL的確還存在一種東西,插件程序~~~

4. c庫函數源碼

不是你表達不清，也許只是你根本不想仔細看一睛VC下面目錄的源碼，事實上就是有的。後附其中的qsort.c，以證明所言不虛。

VC的庫是提供源碼的，這東西也不值錢。
X:\Program Files\Microsoft Visual Studio\VCXX\CRT\SRC
注意有些可能本身是用匯編寫的。

/***
*qsort.c - quicksort algorithm; qsort() library function for sorting arrays
*
* Copyright (c) 1985-1997, Microsoft Corporation. All rights reserved.
*
*Purpose:
* To implement the qsort() routine for sorting arrays.
*
*******************************************************************************/

#include <cruntime.h>
#include <stdlib.h>
#include <search.h>

/* prototypes for local routines */
static void __cdecl shortsort(char *lo, char *hi, unsigned width,
int (__cdecl *comp)(const void *, const void *));
static void __cdecl swap(char *p, char *q, unsigned int width);

/* this parameter defines the cutoff between using quick sort and
insertion sort for arrays; arrays with lengths shorter or equal to the
below value use insertion sort */

#define CUTOFF 8 /* testing shows that this is good value */

/***
*qsort(base, num, wid, comp) - quicksort function for sorting arrays
*
*Purpose:
* quicksort the array of elements
* side effects: sorts in place
*
*Entry:
* char *base = pointer to base of array
* unsigned num = number of elements in the array
* unsigned width = width in bytes of each array element
* int (*comp)() = pointer to function returning analog of strcmp for
* strings, but supplied by user for comparing the array elements.
* it accepts 2 pointers to elements and returns neg if 1<2, 0 if
* 1=2, pos if 1>2.
*
*Exit:
* returns void
*
*Exceptions:
*
*******************************************************************************/

/* sort the array between lo and hi (inclusive) */

void __cdecl qsort (
void *base,
unsigned num,
unsigned width,
int (__cdecl *comp)(const void *, const void *)
)
{
char *lo, *hi; /* ends of sub-array currently sorting */
char *mid; /* points to middle of subarray */
char *loguy, *higuy; /* traveling pointers for partition step */
unsigned size; /* size of the sub-array */
char *lostk[30], *histk[30];
int stkptr; /* stack for saving sub-array to be processed */

/* Note: the number of stack entries required is no more than
1 + log2(size), so 30 is sufficient for any array */

if (num < 2 || width == 0)
return; /* nothing to do */

stkptr = 0; /* initialize stack */

lo = base;
hi = (char *)base + width * (num-1); /* initialize limits */

/* this entry point is for pseudo-recursion calling: setting
lo and hi and jumping to here is like recursion, but stkptr is
prserved, locals aren't, so we preserve stuff on the stack */
recurse:

size = (hi - lo) / width + 1; /* number of el's to sort */

/* below a certain size, it is faster to use a O(n^2) sorting method */
if (size <= CUTOFF) {
shortsort(lo, hi, width, comp);
}
else {
/* First we pick a partititioning element. The efficiency of the
algorithm demands that we find one that is approximately the
median of the values, but also that we select one fast. Using
the first one proces bad performace if the array is already
sorted, so we use the middle one, which would require a very
wierdly arranged array for worst case performance. Testing shows
that a median-of-three algorithm does not, in general, increase
performance. */

mid = lo + (size / 2) * width; /* find middle element */
swap(mid, lo, width); /* swap it to beginning of array */

/* We now wish to partition the array into three pieces, one
consisiting of elements <= partition element, one of elements
equal to the parition element, and one of element >= to it. This
is done below; comments indicate conditions established at every
step. */

loguy = lo;
higuy = hi + width;

/* Note that higuy decreases and loguy increases on every iteration,
so loop must terminate. */
for (;;) {
/* lo <= loguy < hi, lo < higuy <= hi + 1,
A[i] <= A[lo] for lo <= i <= loguy,
A[i] >= A[lo] for higuy <= i <= hi */

do {
loguy += width;
} while (loguy <= hi && comp(loguy, lo) <= 0);

/* lo < loguy <= hi+1, A[i] <= A[lo] for lo <= i < loguy,
either loguy > hi or A[loguy] > A[lo] */

do {
higuy -= width;
} while (higuy > lo && comp(higuy, lo) >= 0);

/* lo-1 <= higuy <= hi, A[i] >= A[lo] for higuy < i <= hi,
either higuy <= lo or A[higuy] < A[lo] */

if (higuy < loguy)
break;

/* if loguy > hi or higuy <= lo, then we would have exited, so
A[loguy] > A[lo], A[higuy] < A[lo],
loguy < hi, highy > lo */

swap(loguy, higuy, width);

/* A[loguy] < A[lo], A[higuy] > A[lo]; so condition at top
of loop is re-established */
}

/* A[i] >= A[lo] for higuy < i <= hi,
A[i] <= A[lo] for lo <= i < loguy,
higuy < loguy, lo <= higuy <= hi
implying:
A[i] >= A[lo] for loguy <= i <= hi,
A[i] <= A[lo] for lo <= i <= higuy,
A[i] = A[lo] for higuy < i < loguy */

swap(lo, higuy, width); /* put partition element in place */

/* OK, now we have the following:
A[i] >= A[higuy] for loguy <= i <= hi,
A[i] <= A[higuy] for lo <= i < higuy
A[i] = A[lo] for higuy <= i < loguy */

/* We've finished the partition, now we want to sort the subarrays
[lo, higuy-1] and [loguy, hi].
We do the smaller one first to minimize stack usage.
We only sort arrays of length 2 or more.*/

if ( higuy - 1 - lo >= hi - loguy ) {
if (lo + width < higuy) {
lostk[stkptr] = lo;
histk[stkptr] = higuy - width;
++stkptr;
} /* save big recursion for later */

if (loguy < hi) {
lo = loguy;
goto recurse; /* do small recursion */
}
}
else {
if (loguy < hi) {
lostk[stkptr] = loguy;
histk[stkptr] = hi;
++stkptr; /* save big recursion for later */
}

if (lo + width < higuy) {
hi = higuy - width;
goto recurse; /* do small recursion */
}
}
}

/* We have sorted the array, except for any pending sorts on the stack.
Check if there are any, and do them. */

--stkptr;
if (stkptr >= 0) {
lo = lostk[stkptr];
hi = histk[stkptr];
goto recurse; /* pop subarray from stack */
}
else
return; /* all subarrays done */
}

/***
*shortsort(hi, lo, width, comp) - insertion sort for sorting short arrays
*
*Purpose:
* sorts the sub-array of elements between lo and hi (inclusive)
* side effects: sorts in place
* assumes that lo < hi
*
*Entry:
* char *lo = pointer to low element to sort
* char *hi = pointer to high element to sort
* unsigned width = width in bytes of each array element
* int (*comp)() = pointer to function returning analog of strcmp for
* strings, but supplied by user for comparing the array elements.
* it accepts 2 pointers to elements and returns neg if 1<2, 0 if
* 1=2, pos if 1>2.
*
*Exit:
* returns void
*
*Exceptions:
*
*******************************************************************************/

static void __cdecl shortsort (
char *lo,
char *hi,
unsigned width,
int (__cdecl *comp)(const void *, const void *)
)
{
char *p, *max;

/* Note: in assertions below, i and j are alway inside original bound of
array to sort. */

while (hi > lo) {
/* A[i] <= A[j] for i <= j, j > hi */
max = lo;
for (p = lo+width; p <= hi; p += width) {
/* A[i] <= A[max] for lo <= i < p */
if (comp(p, max) > 0) {
max = p;
}
/* A[i] <= A[max] for lo <= i <= p */
}

/* A[i] <= A[max] for lo <= i <= hi */

swap(max, hi, width);

/* A[i] <= A[hi] for i <= hi, so A[i] <= A[j] for i <= j, j >= hi */

hi -= width;

/* A[i] <= A[j] for i <= j, j > hi, loop top condition established */
}
/* A[i] <= A[j] for i <= j, j > lo, which implies A[i] <= A[j] for i < j,
so array is sorted */
}

/***
*swap(a, b, width) - swap two elements
*
*Purpose:
* swaps the two array elements of size width
*
*Entry:
* char *a, *b = pointer to two elements to swap
* unsigned width = width in bytes of each array element
*
*Exit:
* returns void
*
*Exceptions:
*
*******************************************************************************/

static void __cdecl swap (
char *a,
char *b,
unsigned width
)
{
char tmp;

if ( a != b )
/* Do the swap one character at a time to avoid potential alignment
problems. */
while ( width-- ) {
tmp = *a;
*a++ = *b;
*b++ = tmp;
}
}

5. 求C語言中的庫函數的源代碼 如printf()函數，我要它的源代碼

如果你安裝的Visual Studio，以及它的Visual C++的話，
那麼在安裝目錄下的VC/crt/src下有所有標准C庫的源代碼

另外，h後綴的頭文件包含函數的聲明，具體的實現都在c後綴的源碼文件中

6. sql server 中如何查看自定義函數的源代碼

如果函數沒有被加密的話(未使用with encrypt子句)，用語句sp_helptext 函數名查看源碼。

如果被加密了，也需要通過第三方工具來解密查看。

使用資料庫引擎創建用於聯機事務處理或聯機分析處理數據的關系資料庫。這包括創建用於存儲數據的表和用於查看、管理和保護數據安全的資料庫對象（如索引、視圖和存儲過程）。可以使用 SQL Server Management Studio 管理資料庫對象，使用 SQL Server Profiler 捕獲伺服器事件。

(6)庫中的函數源碼擴展閱讀

新特性

T-SQL 天生就是基於集合的關系型資料庫管理系統編程語言，可以提供高性能的數據訪問。它與許多新的特性相結合，包括通過同時使用TRY和CTACH來進行錯誤處理，可以在語句中返回一個結果集的通用表表達式，以及通過PIVOT 和UNPIVOT命令將列轉化為行和將列轉化為行的能力。

SQL Server 2005中的第二個主要的增強特性就是整合了符合.NET規范的語言 ，例如C#, 或者是可以構建對象(存儲過程，觸發器，函數等)的VB.NET。