库中的函数源码

1. eclipse中如何用快捷键查找java类库中的源码

这个分两步：

1、鼠标放到你要看的函数上面，按ctrl键，可以看到定义的提示如下图：

2. sql server 中如何查看自定义函数的源代码

可按如下方法查询，以sqlserver2008为例：
1、登录SQL
Server
Management
Studio。
2、展开左边的树，先在数据库中找到自己创建自定义函数的库，如数据库-系统数据库-master。
3、依次点击可编程性-函数-标量值函数，如曾经创建过一个叫“fn_myget”的自定义函数，就能看见。
4、右键此函数，点击编辑，就能看到这个函数的源代码。
5、代码如图，红框部分即为源代码。

3. 如何看c语言标准库函数的源代码

很遗憾,标准库中的函数结合了系统,硬件等的综合能力,是比较近机器的功能实现,所以大部分是用汇编完成的,而且已经导入到了lib和dll里了,就是说,他们已经被编译好了,似乎没有代码的存在了.
能看到的也只有dll中有多少函数被共享.
第三方可能都是dll,因为上面也说了,dll是编译好的,只能看到成品,就可以隐藏代码,保护自己的知识产权,同时也是病毒的归宿...... 当然,除了DLL的确还存在一种东西,插件程序~~~

4. c库函数源码

不是你表达不清，也许只是你根本不想仔细看一睛VC下面目录的源码，事实上就是有的。后附其中的qsort.c，以证明所言不虚。

VC的库是提供源码的，这东西也不值钱。
X:\Program Files\Microsoft Visual Studio\VCXX\CRT\SRC
注意有些可能本身是用汇编写的。

/***
*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语言中的库函数的源代码 如printf()函数，我要它的源代码

如果你安装的Visual Studio，以及它的Visual C++的话，
那么在安装目录下的VC/crt/src下有所有标准C库的源代码

另外，h后缀的头文件包含函数的声明，具体的实现都在c后缀的源码文件中

6. sql server 中如何查看自定义函数的源代码

如果函数没有被加密的话(未使用with encrypt子句)，用语句sp_helptext 函数名查看源码。

如果被加密了，也需要通过第三方工具来解密查看。

使用数据库引擎创建用于联机事务处理或联机分析处理数据的关系数据库。这包括创建用于存储数据的表和用于查看、管理和保护数据安全的数据库对象（如索引、视图和存储过程）。可以使用 SQL Server Management Studio 管理数据库对象，使用 SQL Server Profiler 捕获服务器事件。

(6)库中的函数源码扩展阅读

新特性

T-SQL 天生就是基于集合的关系型数据库管理系统编程语言，可以提供高性能的数据访问。它与许多新的特性相结合，包括通过同时使用TRY和CTACH来进行错误处理，可以在语句中返回一个结果集的通用表表达式，以及通过PIVOT 和UNPIVOT命令将列转化为行和将列转化为行的能力。

SQL Server 2005中的第二个主要的增强特性就是整合了符合.NET规范的语言 ，例如C#, 或者是可以构建对象(存储过程，触发器，函数等)的VB.NET。