无符号整数溢出

Question

Go spec say on unsigned integer overflow:

For unsigned integer values, the operations +, -, *, and << are computed modulo 2n, where n is the bit width of the unsigned integer's type. Loosely speaking, these unsigned integer operations discard high bits upon overflow, and programs may rely on ''wrap around''.

I try to test it, but get inconsistent result - http://play.golang.org/p/sJxtSHbigT:

package main

import "fmt"

func main() {
    fmt.Println("test")
    var num uint32 = 1 << 35
}

This give error:

prog.go:7: constant 34359738368 overflows uint32
 [process exited with non-zero status]

But according to spec should be no error but rather I should seen 0.

Answer 1

The specification you quote refers specifically to the results of "the operations +, -, *, and <<". You're trying to define a constant, not looking at the result of one of those operations.

You also can't use those over-sized values for the input of those operations. The compiler won't wrap any values for you; that's just the runtime behaviour of those operations.

package main

import "fmt"

func main() {
    var num uint32 = 1 + 1 << 35
    fmt.Printf("num = %v
", num)
}

prog.go:6: constant 34359738369 overflows uint32
 [process exited with non-zero status]

---

Here's an interesting example.

var num uint32 = (1 << 31) + (1 << 31)
fmt.Printf("num = %v
", num)

prog.go:6: constant 4294967296 overflows uint32
 [process exited with non-zero status]

In this case, the compiler attempts to evaluate (1 << 31) + (1 << 31) at compile-time, producing the constant value 4294967296, which is too large to fit.

var num uint32 = (1 << 31)
num += (1 << 31)
fmt.Printf("num = %v
", num)

num = 0

In this case, the addition is performed at run-time, and the value wraps around as you'd expect.