如何使用scanf获取指针大小的整数？

Question

我有一些继承的FlexLM代码,它将整数转换为需要在32位和64位机器上工作的指针.使用scanf从程序的参数的argc填充整数以读取整数值.

我应该如何可靠地读取argc字符串以获得适合分配指针的值,以便它可以在32位和64位计算机上运行？

目前代码看起来像这样:

// FlexLM includes this:
typedef char * LM_A_VAL_TYPE; /* so that it will be big enough for */
                              /* any data type on any system */

// My main() includes this:
[...]
if (!strcmp(argv[i], "-maxlen")) {
  int max = 0;
  i++;

  if (i >= argc) {
    break;
  }
  sscanf(argv[i], "%d", &max);
  if (!max) {
    fprintf(stderr, "Error: -maxlen %s Invalid line length\n", argv[i]);
  } else {
    lc_set_attr(lm_job, LM_A_MAX_LICENSE_LEN, (LM_A_VAL_TYPE)max);
  }
}
[...]

起初我以为我可以使用uintptr_t,但我怎么能相应scanf地知道尺寸呢？也许我应该把它作为指针值使用%p,但是手册页让我怀疑它是否可靠地工作:

   p      Matches an implementation-defined set of sequences, which shall be the
          same as the set of sequences that is produced by the %p  conversion
          specification  of  the  corresponding fprintf()  functions.  The
          application  shall  ensure that the corresponding argument is a pointer
          to a pointer to void. The interpretation of the input item is
          implementation-defined. If the input item is a value converted earlier
          during the same program execution, the pointer that results shall
          compare equal to that value; otherwise, the behavior of the %p
          conversion specification is undefined.

我宁愿不使用#ifdef来根据指针大小制作两个单独的版本,因为这对我来说似乎是一个丑陋的代码.

Answer 1

C99 inttypes.h应该定义一个宏SCNuPTR,它是正确的scanf格式说明符,用于平台上的uintptr_t参数.

uintptr_t intptr = 0;
sscanf(argv[i], SCNuPTR, &intptr);