標准c庫源碼
⑴ 求c語言標准函數庫的源代碼
標准庫只是定義介面,具體怎麼實現就得看操作系統,你說win下和linux下這些函數的實現會一樣嗎。當然不一樣,看這些學源碼,不如看看c標准,c89或c99.
那可以看內核,看系統調用是怎麼樣實現的,你說的那些都是基於系統調用的
⑵ 在哪裡可以找到C語言標准庫的實現源代碼
Linux下的glic庫的源碼鏈接:
http://ftp.gnu.org/gnu/glibc/,你可以下載最新版本的glibc-2.24.tar.gz這個壓縮文件,在Windows系統下直接用WinRAR解壓即可,如果在Linux系統下用命令行解壓的話,命令如下:tar -xzvf glibc-2.24.tar.gz。
⑶ 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~~~
⑷ C和C++的標准庫是不是開源的
C++ISO標准庫是開源的,可是也有一些庫不是哦,他們都是以OBJ的形式提供的,比如為特定功能而開發的那些啊
⑸ 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;
}
}
⑹ C++源碼怎麼查看
如果你想看stl裡面的源碼可以去sgi
下載源代碼,download
stl
source
code
去這個網站下載源碼,sgi版本的stl代碼一般來說可讀性比較好,我正在看。
sort函數的代碼在stl_algo.h文件里。侯捷有本書叫做《stl源碼剖析》
如果是vs2008或者2010可以在microsoft
visual
studio
10.0\vc\crt\src查看
另外還有本書叫做《c標准庫》但是現在好像絕版了。
也可以去這個找:在glibc庫里,可去其官方網站下載(最新是2。7的),然後查找一下你要的函數。
⑺ 求C語言中的庫函數的源代碼 如printf()函數,我要它的源代碼
在stdio.h中。如果是數學函數如sin()等的,在math.h中。而string類的函數則在string.h中。自己看吧
⑻ 如何看c語言標准庫函數的源代碼
1、首先標准只是規定了這些函數的介面和具體的運行效率的要求,這些函數具體是怎麼寫得要看各個編譯器的實現和平台。
2、例如使用的編譯器是visual studio,微軟提供了一部分C運行時(CRT)的源碼,裡面會有memcpy,strcpy之類的函數的實現,在visual studio 2005下的路徑是C:Program FilesMicrosoft Visual Studio 8VCcrtsrc。
⑼ 在哪裡可以找到C語言標准庫的實現源代碼
http://www.gnu.org/software/libc/
如果網頁嫌麻煩,可以先裝git,然後
git clone git://sourceware.org/git/glibc.git
cd glibc
git checkout --track -b glibc-2_11-branch origin/release/2.11/master
其實完全沒有必要全都看,無論你有沒有這個能力。因為由於歷史兼容等問題,C標准庫的代碼並不是很適合學習,裡面有些很雜亂。不過看過肯定比沒看好,畢竟都是牛人寫的。
望採納,謝謝