请问C已将做字符串添加任何工具？ [英] Does C have any tools for doing string addition?

问题描述

我在做一个函数，返回重新psented为一棵树$ P $像

I'm making a function that returns the derivative of a function that is represented as a tree like

      /   +    \
     *          ^
   /   \      /   \
  x     5   3.14   x

与形式的节点

with nodes of the form

typedef struct node
{
    char * fx; // function
    struct node * gx; // left-hand side
    char * op; // operator
    struct node * hx; // right-hand side
} node;

如果一个节点没有孩子，如 X ， 5 ， 3.14 在上面的例子中，然后其运， GX 和 HX 是 NULL ，否则它的 FX 是 NULL 。

If a node has no children, e.g. x, 5, 3.14 in the above example, then its op, gx and hx are NULL, and otherwise its fx is NULL.

我的函数计算衍生的样子

My function for calculating the derivative looks like

char * deriveFromTree ( node * rt )
{
    char * buff = malloc(100*sizeof(char));
    int curBuffIdx = 0;
    if (rt->op) // if rt is of the form rt = gx op hx
    {
        char * dgdx = deriveFromTree(rt->gx); // g'(x)
        char * dhdx = deriveFromTree(rt->hx); // h'(x)
        char thisop = *rt->op;
        if (thisop == '+' || thisop == '-')
        {
            // ... want to do equivalent of
            //     buff = dgdx + thisop + dhdx
        }
        else if (thisop == '*')
        {
            // ...
        }
        else if (thisop == '/')
        {
            // ...
        }
        else if (thisop == '^')
        {
            // ...
        }
    }
    else // rt is a base expression -- x or a constant
    {
        buff[curBuffIdx] = strcmp(rt->fx, 'x') ? '1': '0';
    }
    buff[++curBuffIdx] = '\0';
    return buff;
}

但我越来越绊倒上的所有字符串添加。我可以从头开始创建一个字符串加法器，如果有已经在做的一种简洁的方式

but I'm getting tripped up on all the string addition. I could create from scratch a string adder, if there's already a compact way of doing

            // ... want to do equivalent of
            //     buff = dgdx + thisop + dhdx

那么我想使用该工具。

then I'd like to use that tool.

推荐答案

如果您的C标准库是GNU或者* BSD，那么你可能有 asprintf 可用。您可能需要启用一个功能测试宏使用它，但。如果你没有 asprintf 可用，它可以很容易地在C标准定义的 vsnprintf 功能。

If your C standard library is GNU or *BSD, then you probably have asprintf available. You may need to enable a feature test macro to use it, though. If you don't have asprintf available, it can easily be defined in terms of the C standard vsnprintf function.

asprintf 返回格式作为一个新分配的字符串（这是你的责任，免费的结果所以你可以写，例如：

asprintf returns the result of the format as a newly-allocated string (which it is your responsibility to free. So you could write, for example:

char* buff;
int n = asprintf(&buff, "%s%c%s", dgdx, thisop, dhdx);

我通常使用一个包装函数，返回的字符串，而不是长度，所以你可以写：

I usually use a wrapper function, which returns the string rather than the length, so you can write:

char* buff = concatf("%s%c%s", dgdx, thisop, dhdx);

下面是三个简单的实现;第一个将工作系统 vasprintf ;在系统的第二个符合POSIX vsnprintf ;第三个针对Windows，这显然实现了一个不同的的snprintf 接口。

Here are three simple implementations; the first will work on systems with vasprintf; the second on systems with Posix vsnprintf; and the third for Windows, which apparently implements a different snprintf interface.

// Version 1, systems which have vasprintf:
char* concatf(const char* fmt, ...) {
  va_list args;
  char* buf = NULL;
  va_start(args, fmt);
  int n = vasprintf(&buf, fmt, args);
  va_end(args);
  if (n < 0) { free(buf); buf = NULL; }
  return buf;
}

// Version 2: Systems without vasprintf but with vsnprintf
char* concatf(const char* fmt, ...) {
  va_list args;
  va_start(args, fmt);
  char* buf = NULL;
  int n = vsnprintf(NULL, 0, fmt, args);
  va_end(args);
  if (n >= 0) {
    va_start(args, fmt);
    buf = malloc(n+1);
    if (buf) vsnprintf(buf, n+1, fmt, args);
    va_end(args);
  }
  return buf;
}

// Version 3: Windows
// Apparently, the implementation of vsnprintf on Windows returns -1
// if not enough space has been provided. So here is the above code
// rewritten according to the documentation I found in
//  https://msdn.microsoft.com/en-us/library/w05tbk72%28VS.71%29.aspx
// and
//  https://msdn.microsoft.com/en-us/library/1kt27hek%28v=vs.71%29.aspx
// but totally untested. (If you try it, let me know)
char* concatf(const char* fmt, ...) {
  char* buf = NULL;
  va_list args;
  va_start(args, fmt);
  int n = _vscprintf(fmt, args);
  va_end(args);
  if (n >= 0) {
    va_start(args, fmt);
    buf = malloc(n+1);
    if (buf) _vsnprintf(buf, n+1, fmt, args);
    va_end(args);
  }
  return buf;
}

这是最简洁的等效我所知道的字符串连接运营商在其他语言。 （这并不一定是最高效的执行时间，但它可能是程序员的时间。）

That's the most concise equivalent I know of to string concatenation operators in other languages. (It's not necessarily the most efficient in execution time, but it probably is in programmer time.)

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