unsigned int u = 1234;
int i = -5678;

unsigned int result = u + i;

What implicit conversions are going on here, and is this code safe for all values of u and i? (Safe, in the sense that even though result in this example will overflow to some huge positive number, I could cast it back to an int and get the real result.)

Answer:

Short Answer

Your i will be converted to an unsigned integer by adding UINT_MAX + 1, then the addition will be carried out with the unsigned values, resulting in a large result (depending on the values of u andi).

Long Answer

According to the C99 Standard:

6.3.1.8 Usual arithmetic conversions

1. If both operands have the same type, then no further conversion is needed.
2. Otherwise, if both operands have signed integer types or both have unsigned integer types, the operand with the type of lesser integer conversion rank is converted to the type of the operand with greater rank.
3. Otherwise, if the operand that has unsigned integer type has rank greater or equal to the rank of the type of the other operand, then the operand with signed integer type is converted to the type of the operand with unsigned integer type.
4. Otherwise, if the type of the operand with signed integer type can represent all of the values of the type of the operand with unsigned integer type, then the operand with unsigned integer type is converted to the type of the operand with signed integer type.
5. Otherwise, both operands are converted to the unsigned integer type corresponding to the type of the operand with signed integer type.

In your case, we have one unsigned int (u) and signed int (i). Referring to (3) above, since both operands have the same rank, your i will need to be converted to an unsigned integer.

6.3.1.3 Signed and unsigned integers

1. When a value with integer type is converted to another integer type other than _Bool, if the value can be represented by the new type, it is unchanged.
2. Otherwise, if the new type is unsigned, the value is converted by repeatedly adding or subtracting one more than the maximum value that can be represented in the new type until the value is in the range of the new type.
3. Otherwise, the new type is signed and the value cannot be represented in it; either the result is implementation-defined or an implementation-defined signal is raised.

Now we need to refer to (2) above. Your i will be converted to an unsigned value by adding UINT_MAX + 1. So the result will depend on how UINT_MAX is defined on your implementation. It will be large, but it will not overflow, because:

6.2.5 (9)

A computation involving unsigned operands can never overflow, because a result that cannot be represented by the resulting unsigned integer type is reduced modulo the number that is one greater than the largest value that can be represented by the resulting type.

Bonus: Arithmetic Conversion Semi-WTF

#include <stdio.h>

int main(void)
{
  unsigned int plus_one = 1;
  int minus_one = -1;

  if(plus_one < minus_one)
    printf("1 < -1");
  else
    printf("boring");

  return 0;
}

You can use this link to try this online: http://codepad.org/yPhYCMFO

Bonus: Arithmetic Conversion Side Effect

Arithmetic conversion rules can be used to get the value of UINT_MAX by initializing an unsigned value to -1, ie:

unsigned int umax = -1; // umax set to UINT_MAX

This is guaranteed to be portable regardless of the signed number representation of the system because of the conversion rules described above. See this SO question for more information: Is it safe to use -1 to set all bits to true?

所有的无符号运算都是以2的n次方为模。如果算术运算符的一个操作数是有符号书，另一个是无符号数，那么有符号数

会被转换为无符号数(表示范围小的总是被转换为表示范围大的)，那么溢出也不会发生。但是，当两个操作数都是有符号数

时，溢出就有可能发生。而且溢出的结果是未定义的。当一个运算的结果发生溢出时，任何假设都是不安全的。

例如，假定a和b是两个非负的整型变量(有符号)，我们需要检查a+b是否溢出，一种想当然的方式是:

if (a + b < 0)

      溢出；

实际上，在现实世界里，这并不能正常运行。当a+b确实发生溢出时，所有关于结果如何的假设均不可靠。比如，在某些

机器的cpu，加法运算将设置一个内部寄存器为四种状态:正，负，零和溢出。在这种机器上，c编译器完全有理由实现以上

的例子，使得a+b返回的不是负，而是这个内存寄存器的溢出状态。显然，if的判断会失败。

一种正确的方式是将a和b都强制转换为无符号整数：

if ( (unsigned)a + (unsigned)b  > INT_MAX)

      溢出;

这里的int_max值为有符号整型的最大值。在一般的编译器里是一个预定义的常量。ANSI C在limits里定义了INT_MAX，值为

2的31次方-1.

不需要用到无符号算数运算的另一种可行方法是：

if (a > INT_MAX - b )

     溢出；