下载c语言库函数源码 c语言库函数大全下载
C语言库函数qsort源代码
void __fileDECL qsort (
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void *base,
size_t num,
size_t width,
int (__fileDECL *comp)(const void *, const void *)
)
#endif /* __USE_CONTEXT */
{
char *lo, *hi; /* ends of sub-array currently sorting */
char *mid; /* points to middle of subarray */
char *loguy, *higuy; /* traveling pointers for partition step */
size_t size; /* size of the sub-array */
char *lostk[STKSIZ], *histk[STKSIZ];
int stkptr; /* stack for saving sub-array to be processed */
/* validation section */
_VALIDATE_RETURN_VOID(base != NULL || num == 0, EINVAL);
_VALIDATE_RETURN_VOID(width 0, EINVAL);
_VALIDATE_RETURN_VOID(comp != NULL, EINVAL);
if (num 2)
return; /* nothing to do */
stkptr = 0; /* initialize stack */
lo = (char *)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
preserved, 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, context);
}
else {
/* First we pick a partitioning 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. We choose the
median of the first, middle, and last elements, to avoid bad
performance in the face of already sorted data, or data that is made
up of multiple sorted runs appended together. Testing shows that a
median-of-three algorithm provides better performance than simply
picking the middle element for the latter case. */
mid = lo + (size / 2) * width; /* find middle element */
/* Sort the first, middle, last elements into order */
if (__COMPARE(context, lo, mid) 0) {
swap(lo, mid, width);
}
if (__COMPARE(context, lo, hi) 0) {
swap(lo, hi, width);
}
if (__COMPARE(context, mid, hi) 0) {
swap(mid, hi, width);
}
/* We now wish to partition the array into three pieces, one consisting
of elements = partition element, one of elements equal to the
partition element, and one of elements than it. This is done
below; comments indicate conditions established at every step. */
loguy = lo;
higuy = hi;
/* Note that higuy decreases and loguy increases on every iteration,
so loop must terminate. */
for (;;) {
/* lo = loguy hi, lo higuy = hi,
A[i] = A[mid] for lo = i = loguy,
A[i] A[mid] for higuy = i hi,
A[hi] = A[mid] */
/* The doubled loop is to avoid calling comp(mid,mid), since some
existing comparison funcs don't work when passed the same
value for both pointers. */
if (mid loguy) {
do {
loguy += width;
} while (loguy mid __COMPARE(context, loguy, mid) = 0);
}
if (mid = loguy) {
do {
loguy += width;
} while (loguy = hi __COMPARE(context, loguy, mid) = 0);
}
/* lo loguy = hi+1, A[i] = A[mid] for lo = i loguy,
either loguy hi or A[loguy] A[mid] */
do {
higuy -= width;
} while (higuy mid __COMPARE(context, higuy, mid) 0);
/* lo = higuy hi, A[i] A[mid] for higuy i hi,
either higuy == lo or A[higuy] = A[mid] */
if (higuy loguy)
break;
/* if loguy hi or higuy == lo, then we would have exited, so
A[loguy] A[mid], A[higuy] = A[mid],
loguy = hi, higuy lo */
swap(loguy, higuy, width);
/* If the partition element was moved, follow it. Only need
to check for mid == higuy, since before the swap,
A[loguy] A[mid] implies loguy != mid. */
if (mid == higuy)
mid = loguy;
/* A[loguy] = A[mid], A[higuy] A[mid]; so condition at top
of loop is re-established */
}
/* A[i] = A[mid] for lo = i loguy,
A[i] A[mid] for higuy i hi,
A[hi] = A[mid]
higuy loguy
implying:
higuy == loguy-1
or higuy == hi - 1, loguy == hi + 1, A[hi] == A[mid] */
/* Find adjacent elements equal to the partition element. The
doubled loop is to avoid calling comp(mid,mid), since some
existing comparison funcs don't work when passed the same value
for both pointers. */
higuy += width;
if (mid higuy) {
do {
higuy -= width;
} while (higuy mid __COMPARE(context, higuy, mid) == 0);
}
if (mid = higuy) {
do {
higuy -= width;
} while (higuy lo __COMPARE(context, higuy, mid) == 0);
}
/* OK, now we have the following:
higuy loguy
lo = higuy = hi
A[i] = A[mid] for lo = i = higuy
A[i] == A[mid] for higuy i loguy
A[i] A[mid] for loguy = i hi
A[hi] = A[mid] */
/* We've finished the partition, now we want to sort the subarrays
[lo, higuy] and [loguy, hi].
We do the smaller one first to minimize stack usage.
We only sort arrays of length 2 or more.*/
if ( higuy - lo = hi - loguy ) {
if (lo higuy) {
lostk[stkptr] = lo;
histk[stkptr] = higuy;
++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 higuy) {
hi = higuy;
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 */
}
求C语言标准函数库的源代码
标准库只是定义接口，具体怎么实现就得看操作系统，你说win下和linux下这些函数的实现会一样吗。当然不一样，看这些学源码，不如看看c标准，c89或c99.
那可以看内核，看系统调用是怎么样实现的，你说的那些都是基于系统调用的
C语言库函数源代码在哪里有看
有安装vs2008或2010吗，在安装目录下面的VC/src中自带有源代码。比如我的就在
D:\Program Files\Microsoft Visual Studio 10.0\VC\crt\src中。没有的话发给你
在哪里可以找到C语言标准库的实现源代码
如果网页嫌麻烦，可以先装git，然后
git clone git://sourceware.org/git/glibc.git
cd glibc
git checkout --track -b glibc-2_11-branch origin/release/2.11/master
其实完全没有必要全都看，无论你有没有这个能力。因为由于历史兼容等问题，C标准库的代码并不是很适合学习，里面有些很杂乱。不过看过肯定比没看好，毕竟都是牛人写的。
望采纳，谢谢
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
如何看c语言标准库函数的源代码?
1、首先标准只是规定了这些函数的接口和具体的运行效率的要求，这些函数具体是怎么写得要看各个编译器的实现和平台。
2、例如使用的编译器是visual studio，微软提供了一部分C运行时(CRT)的源码，里面会有memcpy，strcpy之类的函数的实现，在visual studio 2005下的路径是C:\Program Files\Microsoft Visual Studio 8\VC\crt\src。
C语言
C语言是一门通用计算机编程语言，应用广泛。C语言的设计目标是提供一种能以简易的方式编译、处理低级存储器、产生少量的机器码以及不需要任何运行环境支持便能运行的编程语言。
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