C语言 如何将pid_t转换为字符串
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How to convert pid_t to string
提问by btevfik
So, I need to strcat pid to some string. I have this
所以，我需要将 pid 转换为某个字符串。我有这个
strcat (str,(char*)getpid());
but this doesn't work.
但这不起作用。
----edit----
- - 编辑 - -
ok i understand the downvotes. i was too quick to post a question. and didnt realize pid returns an int and i can't cast an int to char*
好的，我理解downvotes。我太快发表问题了。并没有意识到 pid 返回一个 int 并且我不能将一个 int 转换为 char*
itoa doesnt work because it is not standart c.
itoa 不起作用，因为它不是标准 c。
this is how i did it.
我就是这样做的。
char pid[10];
snprintf(pid, 10,"%d",(int)getpid());
strcat (str, pid);
回答by nneonneo
Instead of using strcatto build a string, consider using the much more flexible (and efficient!) sprintf/snprintffunction instead:
与其使用strcat构建字符串，不如考虑使用更灵活（更高效！）的sprintf/snprintf函数：
char *end = str;
end += sprintf(end, "%s ", "hello!");
end += sprintf(end, "%ld", (long)getpid());
if(bar)
end += sprintf(end, "%x", 0xf00d);
Observe that sprintfreturns the number of characters written, so you can build a string without succumbing to Schlemiel the Painter's algorithm. If you want to ensure that you don't overrun a buffer of fixed size, snprintfwill do this for you (strcatwill just blindly concatenate, and there is no standard way to avoid that).
观察sprintf返回写入的字符数，因此您可以构建字符串而不会屈服于Schlemiel the Painter's algorithm。如果您想确保不会超出固定大小的缓冲区，snprintf将为您执行此操作（strcat只会盲目连接，并且没有标准方法可以避免这种情况）。
Note that the pid_tstandardguarantees that there are "one or more programming environments in which the [width] of pid_t... is no greater than the width of type long". Therefore, casting pid_tto longis safe as long as getconfsays so.
请注意，该pid_t标准保证存在“一个或多个编程环境，其中 ... 的 [宽度]pid_t不大于类型的宽度long”。因此，只要这么说，投射pid_t到long是安全的getconf。
回答by autistic
I see several references that mention casting to long, and then using the %ldformat specifier of sprintf: sprintf(str + strlen(str), "%ld", (long) getpid());
我看到几个参考文献提到转换为 long，然后使用%ldsprintf的格式说明符：sprintf(str + strlen(str), "%ld", (long) getpid());
