如何查看C语言，内库的源代码？
如果是“.cpp”文件并且有VC++的环境，可直接双击文件打开或者先打开编译环境，在新建一个控制台下的源文件，然后，选择file菜单下的open找到你的文件导入，然后编译运行；如果是其他格式的，如txt文件，也可先打开编译环境，新建一个控制台下的源文件，然后直接复制粘贴进去，然后编译运行；
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便已运行的操作如图：
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
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~~~
在哪里可以找到C语言标准库的实现源代码
Linux下的glic库的源码链接：
，你可以下载最新版本的glibc-2.24.tar.gz这个压缩文件，在Windows系统下直接用WinRAR解压即可，如果在Linux系统下用命令行解压的话，命令如下：tar -xzvf glibc-2.24.tar.gz。
请问到哪里可以找C语言的库函数的代码,例如PRINTF函数的代码
这个你找不到,我曾经在学习过程中也找过,只能在include文件夹下找到对于函数的定义,但函数体部分不可见.你可以到LINUX系统下找找,LINUX是完全开放源代码的.
C语言库函数qsort源代码
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 */
}
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