C++ left shift operation

Question

Here is my code

int a=2147483647; 
int b= a<<1;
cout<<"a="<<a<<",  b="<<b;

The output I am getting is- a=214783647, b=-2

Binary representation of a is

0111 1111 1111 1111 1111 1111 1111 1111

By shifting it by 1 bit, it will change sign bit and replace LSB with 0. So, I think answer will be -ve and magnitude will be subtracted by 1 ie -2147483646 But it is giving result as -2 . Please explain.

Answer 1

[expr.shift]/1 The value of E1 << E2 is E1 left-shifted E2 bit positions; vacated bits are zero-filled. ... if E1 has a signed type and non-negative value, and E1 × 2^E2 is representable in the corresponding unsigned type of the result type, then that value, converted to the result type, is the resulting value; otherwise, the behavior is undefined .

Emphasis mine. Your program exhibits undefined behavior.

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Edit: Upon closer consideration, I no longer think it's undefined behavior. 2147483647*2 does fit into unsigned int , "the corresponding unsigned type" of int . Its conversion to int is not undefined, but merely implementation-defined. It's entirely reasonable for an implementation using two's complement to define this conversion so that 2147483647*2 == -2 , just reinterpreting the bit pattern, as other answers explained.

Answer 2

This is because your computer is using 2 complement for the signed value. Unsigned shifted value is 0xFFFFFFFE , which is -2 in 2 complement, not -2147483647 .

Shifting is implementation defined in C.

BTW, -2147483647 is 0x80000001 on such CPU.

Answer 3

Well, there is a very long story behind.

Since int is a signed type, it means that the first bit is a sign and the whole system is two-complement.

so x = 0b 1111 1111 1111 1111 1111 1111 1111 0111 is x = -9 and for example x = 0b 1111 1111 1111 1111 1111 1111 1111 1111 is x = -1 and x = 0b 0000 0000 0000 0000 0000 0000 0000 0010 is 2

Learn more about Two complement .