c语言 - size_t溢出时，"<"和">"运算符是否正常工作？

Question

我正在研究一个项目以更新20岁的代码，许多问题与整数溢出有关。我想确保正确测试溢出，所以我编写了一个测试程序。输出使我鼓舞。在这里是：

#include <stdlib.h>
#include <stdio.h>
#include <string.h>
#include <limits.h>
int main (void) {
   size_t largerNum,Num;
   largerNum = 12;
   Num = UINT_MAX;
   printf("largerNum = %unNum = %unNum + 1 = %un", largerNum    , Num, Num + 1);
   largerNum = Num + 1;
   printf("largerNum now = %un", largerNum);
   if(largerNum < Num ){
      printf("largerNum overflowed to %un", largerNum);
   }
   else {
      printf("largerNum did not overflow: %un", largerNum);
   }
   printf("Is (0 < UINT_MAX)?n");
   (0 < UINT_MAX)?printf("YESn"):printf("NOn");
   printf("Is (largerNum < Num)?n");
   (largerNum < Num)?printf("YESn"):printf("NOn");
   return 0;
}

及其输出：

[afischer@susm603 /home/afischer/Fischer_Playground/overflowTest]$ main
largerNum = 12
Num = 4294967295
Num + 1 = 0
largerNum now = 0
largerNum did not overflow: 0
Is (0 < UINT_MAX)?
YES
Is (largerNum < Num)?
NO

我在这里和这里查看了其他一些帖子，并阅读了本文，但是它尚未使输出清楚地表明。有人以前看到了吗？

编辑：从size_t更改为unsigned long时，我让它起作用，这不应该做任何事情。

  6 int main (void) {
  7 
  8    unsigned long largerNum,Num;
  9 
 10    largerNum = 12;
 11    Num = UINT_MAX;
 12 
 13    printf("largerNum = %unNum = %unNum + 1 = %un", largerNum    , Num, Num + 1);
 14 
 15    largerNum = Num + 2;
 16 
 17    printf("largerNum now = %un", largerNum);
 18 
 19    if(largerNum < Num ){
 20       printf("largerNum overflowed to %un", largerNum);
 21    }
 22    else {
 23       printf("largerNum did not overflow: %un", largerNum);
 24    }
 25 
 26    printf("Is (0 < UINT_MAX)?n");
 27 
 28    (0 < UINT_MAX)?printf("YESn"):printf("NOn");
 29 
 30    printf("Is (largerNum < Num)?n");
 31 
 32    (largerNum < Num)?printf("YESn"):printf("NOn");
 33 
 34 
 35    printf("largerNum = %un", largerNum);
 36    printf("Num = %un", Num);
 37 
 38    return 0;
 39 }

输出：

[afischer@susm603 /home/afischer/Fischer_Playground/overflowTest]$ main
largerNum = 12
Num = 4294967295
Num + 1 = 0
largerNum now = 1
largerNum overflowed to 1
Is (0 < UINT_MAX)?
YES
Is (largerNum < Num)?
YES
largerNum = 1
Num = 4294967295

edit2：

阅读一些评论后，我将" uint_max"替换为" ulong_max"，而三元运营商正常运行。然后，我将" size_t"更改为"未签名长"。而且它仍然可以正常工作。对我来说奇怪的是，在我的机器上，" size_t"，" unsigned int"one_answers" unsigned long"都是相同数量的字节，而'uint_max'和'ulong_max'是相同的值尽管一切都一样，但仍然会失败。也许不一样？这使我对c。

的理解感到不安

对于那些有兴趣的人，工作代码：

  6 int main (void) {
  7    /* Can be size_t or unsigned long */
  8    size_t largerNum,Num;
  9 
 10    largerNum = 12;
 11    Num = ULONG_MAX;
 12 
 13    printf("largerNum = %unNum = %unNum + 1 = %un", largerNum    , Num, Num + 1);
 14 
 15    largerNum = Num + 2;
 16 
 17    printf("largerNum now = %un", largerNum);
 18 
 19    if(largerNum < Num ){
 20       printf("largerNum overflowed to %un", largerNum);
 21    }
 22    else {
 23       printf("largerNum did not overflow: %un", largerNum);
 24    }
 25 
 26    printf("Is (0 < ULONG_MAX)?n");
 27 
 28    (0 < ULONG_MAX)?printf("YESn"):printf("NOn");
 29 
 30    printf("Is (largerNum < Num)?n");
 31 
 32    (largerNum < Num)?printf("YESn"):printf("NOn");
 33 
 34    
 35    printf("largerNum = %un", largerNum);
 36    printf("Num = %un", Num);
 37    
 38    return 0;
 39 }

输出：

[afischer@susm603 /home/afischer/Fischer_Playground/overflowTest]$ main
largerNum = 12
Num = 4294967295
Num + 1 = 0
largerNum now = 1
largerNum overflowed to 1
Is (0 < ULONG_MAX)?
YES
Is (largerNum < Num)?
YES
largerNum = 1
Num = 4294967295

最终编辑：

阅读了更多评论后，我发现我的printf()语句错误。谢谢大家的帮助，现在一切都更有意义。= D

最终代码：

  6 int main (void) {
  7 
  8    unsigned long largerNum,Num;
  9 
 10    largerNum = 12;
 11    Num = ULONG_MAX;
 12 
 13    printf("largerNum = %zunNum = %zunNum + 1 = %zun", larger    Num, Num, Num + 1);
 14 
 15    largerNum = Num + 2;
 16 
 17    printf("largerNum now = %zun", largerNum);
 18 
 19    if(largerNum < Num ){
 20       printf("largerNum overflowed to %zun", largerNum);
 21    }
 22    else {
 23       printf("largerNum did not overflow: %zun", largerNum);
 24    }
 25 
 26    printf("Is (0 < ULONG_MAX)?n");
 27 
 28    (0 < ULONG_MAX)?printf("YESn"):printf("NOn");
 29 
 30    printf("Is (largerNum < Num)?n");
 31 
 32    (largerNum < Num)?printf("YESn"):printf("NOn");
 33 
 34 
 35    printf("largerNum = %zun", largerNum);
 36    printf("Num = %zun", Num);
 37 
 38    return 0;
 39 }

最终输出：

[afischer@susm603 /home/afischer/Fischer_Playground/overflowTest]$ main
largerNum = 12
Num = 18446744073709551615
Num + 1 = 0
largerNum now = 1
largerNum overflowed to 1
Is (0 < ULONG_MAX)?
YES
Is (largerNum < Num)?
YES
largerNum = 1
Num = 18446744073709551615

Answer 1

我的猜测是您的平台具有64位的size_t，并且您使用错误的格式指定符打印size_t，这是未定义的行为，并导致误导性输出。

要打印size_t s，在GCC和Clang上使用%zu，在MSVC上使用%Iu。或者忘记所有这些，然后使用std::cout打印结果。

使用VS2015上的%Iu，我在64位编译器上获得的输出是

largerNum = 12
Num = 4294967295
Num + 1 = 4294967296
largerNum now = 4294967296
largerNum did not overflow: 4294967296
Is (0 < UINT_MAX)?
YES
Is (largerNum < Num)?
NO

Answer 2

只是添加到 @praetorian的答案中并显示一种安全实现：

#include <iostream>
#include <limits>
int main (void) {
   using std::size_t;
   using std::cout;
   size_t largerNum = 12;
   size_t Num = std::numeric_limits<size_t>::max();
   cout << "largerNum = " << largerNum << "nNum = " << Num << "nNum + 1 = " << Num + 1 << "n";
   largerNum = Num + 1;
   cout << "largerNum now = " << largerNum << "n";
   if(largerNum < Num ){
       cout << "largerNum overflowed to " << largerNum << "n";
   }
   else {
       cout << "largerNum did not overflow: " << largerNum << "n";
   }
   cout << "Is (0 < Unsigned Maximum)?n";
   (0 < std::numeric_limits<size_t>::max())?cout << "YESn":cout << "NOn";
   cout << "Is (largerNum < Num)?n";
   (largerNum < Num)?cout << "YESn":cout << "NOn";
   return 0;
}

因此：'printf'在C 中不是一个不错的选择，它不是类型的安全性（尽管好的编译器可以识别无效的格式指定器）。另一方面，iostream运算符很麻烦（对于许多人来说），如果必须将输出转换为不同的语言（例如：gnu getline），则很糟糕。您可以浏览网络以获取类型的安全格式字符串（例如：boost ::格式）