如何查看C語言，內庫的源代碼？

如果是“.cpp”文件并且有VC++的環境，可直接雙擊文件打開或者先打開編譯環境，在新建一個控制臺下的源文件，然后，選擇file菜單下的open找到你的文件導入，然后編譯運行；如果是其他格式的，如txt文件，也可先打開編譯環境，新建一個控制臺下的源文件，然后直接復制粘貼進去，然后編譯運行；

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便已運行的操作如圖：

求C語言標準函數庫的源代碼

標準庫只是定義接口，具體怎么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標準，c89或c99.

那可以看內核，看系統調用是怎么樣實現的，你說的那些都是基于系統調用的

如何看c語言標準庫函數的源代碼?

很遺憾,標準庫中的函數結合了系統,硬件等的綜合能力,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~~~

在哪里可以找到C語言標準庫的實現源代碼

Linux下的glic庫的源碼鏈接：

，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。

請問到哪里可以找C語言的庫函數的代碼,例如PRINTF函數的代碼

這個你找不到,我曾經在學習過程中也找過,只能在include文件夾下找到對于函數的定義,但函數體部分不可見.你可以到LINUX系統下找找,LINUX是完全開放源代碼的.

C語言庫函數qsort源代碼

void __fileDECL qsort (

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 */

}