当 size_t 溢出时，“<”和“>”运算符是否正常工作？

Question

我正在做一个更新 20 年前代码的项目，其中很多问题都与整数溢出有关。我想确保我正确地测试了溢出，所以我写了一个测试程序。它的输出让我吃惊。这是：

#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 = %u\nNum = %u\nNum + 1 = %u\n", largerNum    , Num, Num + 1);

   largerNum = Num + 1;

   printf("largerNum now = %u\n", largerNum);

   if(largerNum < Num ){
      printf("largerNum overflowed to %u\n", largerNum);
   }
   else {
      printf("largerNum did not overflow: %u\n", largerNum);
   }

   printf("Is (0 < UINT_MAX)?\n");

   (0 < UINT_MAX)?printf("YES\n"):printf("NO\n");

   printf("Is (largerNum < Num)?\n");

   (largerNum < Num)?printf("YES\n"):printf("NO\n");

   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

我查看了其他一些帖子 here and here and read this paper，但它并没有使输出更清晰。有人以前看过这个吗？

编辑：我在从 size_t 更改为 unsigned long 时开始工作，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 = %u\nNum = %u\nNum + 1 = %u\n", largerNum    , Num, Num + 1);
 14 
 15    largerNum = Num + 2;
 16 
 17    printf("largerNum now = %u\n", largerNum);
 18 
 19    if(largerNum < Num ){
 20       printf("largerNum overflowed to %u\n", largerNum);
 21    }
 22    else {
 23       printf("largerNum did not overflow: %u\n", largerNum);
 24    }
 25 
 26    printf("Is (0 < UINT_MAX)?\n");
 27 
 28    (0 < UINT_MAX)?printf("YES\n"):printf("NO\n");
 29 
 30    printf("Is (largerNum < Num)?\n");
 31 
 32    (largerNum < Num)?printf("YES\n"):printf("NO\n");
 33 
 34 
 35    printf("largerNum = %u\n", largerNum);
 36    printf("Num = %u\n", 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

编辑2：

阅读一些评论后，我将 'UINT_MAX' 替换为 'ULONG_MAX'，三元运算符运行正常。然后我将 'size_t' 更改为 'unsigned long'。它仍然可以正常工作。令我感到奇怪的是，在我的机器上，'size_t'、'unsigned int' 和 'unsigned long' 都是相同的字节数，而 'UINT_MAX' 和 'ULONG_MAX' 是相同的值，但尽管一切都相同，但该三元运算符仍然会失败。也许不一样？这打乱了我对 C 的理解。

感兴趣的朋友，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 = %u\nNum = %u\nNum + 1 = %u\n", largerNum    , Num, Num + 1);
 14 
 15    largerNum = Num + 2;
 16 
 17    printf("largerNum now = %u\n", largerNum);
 18 
 19    if(largerNum < Num ){
 20       printf("largerNum overflowed to %u\n", largerNum);
 21    }
 22    else {
 23       printf("largerNum did not overflow: %u\n", largerNum);
 24    }
 25 
 26    printf("Is (0 < ULONG_MAX)?\n");
 27 
 28    (0 < ULONG_MAX)?printf("YES\n"):printf("NO\n");
 29 
 30    printf("Is (largerNum < Num)?\n");
 31 
 32    (largerNum < Num)?printf("YES\n"):printf("NO\n");
 33 
 34    
 35    printf("largerNum = %u\n", largerNum);
 36    printf("Num = %u\n", 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：

阅读更多评论后，我发现我的 printf() 陈述是错误的。谢谢大家的帮助，现在一切都变得更有意义了。 =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 = %zu\nNum = %zu\nNum + 1 = %zu\n", larger    Num, Num, Num + 1);
 14 
 15    largerNum = Num + 2;
 16 
 17    printf("largerNum now = %zu\n", largerNum);
 18 
 19    if(largerNum < Num ){
 20       printf("largerNum overflowed to %zu\n", largerNum);
 21    }
 22    else {
 23       printf("largerNum did not overflow: %zu\n", largerNum);
 24    }
 25 
 26    printf("Is (0 < ULONG_MAX)?\n");
 27 
 28    (0 < ULONG_MAX)?printf("YES\n"):printf("NO\n");
 29 
 30    printf("Is (largerNum < Num)?\n");
 31 
 32    (largerNum < Num)?printf("YES\n"):printf("NO\n");
 33 
 34 
 35    printf("largerNum = %zu\n", largerNum);
 36    printf("Num = %zu\n", 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

我猜你的平台有 64 位 size_t，你使用错误的格式说明符来打印 size_t，这是 undefined behavior 并导致误导性的输出。

至 print size_ts，在 gcc 和 clang 上使用 %zu，在 MSVC 上使用 %Iu。或者忘记所有这些并使用 std::cout 打印结果。

在 VS2015 上使用 %Iu，我在 64 位编译器上得到的输出是

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

只需添加到@Praetorian 的答案并显示类型安全的实现：

#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 << "YES\n":cout << "NO\n";

   cout << "Is (largerNum < Num)?\n";

   (largerNum < Num)?cout << "YES\n":cout << "NO\n";

   return 0;
}

因此：'printf' 在 C++ 中不是一个好的选择，它不是类型安全的（尽管好的编译器可以识别无效的格式说明符）。另一方面，iostream 运算符很麻烦（对许多人来说），如果必须将输出翻译成不同的语言（例如：gnu getline），则效果不佳。您可以在网上浏览类型安全的格式字符串（例如：boost::format）

当 size_t 溢出时，“<”和“>”运算符是否正常工作？

Do the "<" and ">" operators work correctly when size_t overflow?

c

printf

overflow

size-t

operator-keyword