C語(yǔ)言庫(kù)函數(shù)qsort源代碼

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 */

}

如何查看C語(yǔ)言，內(nèi)庫(kù)的源代碼？

如果是“.cpp”文件并且有VC++的環(huán)境，可直接雙擊文件打開或者先打開編譯環(huán)境，在新建一個(gè)控制臺(tái)下的源文件，然后，選擇file菜單下的open找到你的文件導(dǎo)入，然后編譯運(yùn)行；如果是其他格式的，如txt文件，也可先打開編譯環(huán)境，新建一個(gè)控制臺(tái)下的源文件，然后直接復(fù)制粘貼進(jìn)去，然后編譯運(yùn)行；

便已運(yùn)行的操作如圖：

如何看c語(yǔ)言標(biāo)準(zhǔn)庫(kù)函數(shù)的源代碼?

1、首先標(biāo)準(zhǔn)只是規(guī)定了這些函數(shù)的接口和具體的運(yùn)行效率的要求，這些函數(shù)具體是怎么寫得要看各個(gè)編譯器的實(shí)現(xiàn)和平臺(tái)。

2、例如使用的編譯器是visual studio，微軟提供了一部分C運(yùn)行時(shí)(CRT)的源碼，里面會(huì)有memcpy，strcpy之類的函數(shù)的實(shí)現(xiàn)，在visual studio 2005下的路徑是C:\Program Files\Microsoft Visual Studio 8\VC\crt\src。

C語(yǔ)言

C語(yǔ)言是一門通用計(jì)算機(jī)編程語(yǔ)言，應(yīng)用廣泛。C語(yǔ)言的設(shè)計(jì)目標(biāo)是提供一種能以簡(jiǎn)易的方式編譯、處理低級(jí)存儲(chǔ)器、產(chǎn)生少量的機(jī)器碼以及不需要任何運(yùn)行環(huán)境支持便能運(yùn)行的編程語(yǔ)言。

本文標(biāo)題：c語(yǔ)言庫(kù)函數(shù)原碼 c語(yǔ)言庫(kù)函數(shù)math
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