CODEWITHSUNDEEP
c
Gillian
Gillian

Reputation: 67

c language interpretation, involves integer calculation

I have the following code, 

void main(void) {
    char x = 1 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 - 1;
    printf("%d", x);
}

I get the result -1

However, 1 * 2 * 2 * 2 * 2 * 2 * 2 * 2, already gets -128. Can you explain why I can successfully run the code above and get the result -1?

Also, is this an example of integer negation or overflow?

Upvotes: 2

Views: 67

Answers (1)

chux
chux

Reputation: 154272

1 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 is int math with a product of 256.
1 * 2 * 2 * 2 * 2 * 2 * 2 * 2 * 2 - 1 is 255.

So code is like 

char x = 255;
printf("%d", x);

If a char is an unsigned char with the range of 0...255, the output is

 255

If a char is a signed char with the range of -128...127, the assignment converts the 255 in an implementation defined manner to a char - likely a wrap around1 (or 255 - 256) with a value of -1. The output is

-1

char is implemented as an unsigned char or signed char.
In OP's case, it is a signed char.
A signed char has the same range, size, encoding as signed char, yet is a distinct type.

is this an example of integer negation or overflow?

Neither. It is an example of an implementation defined conversion. And a very common one at that.

Conversions .... 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. C17dr § 6.3.1.3 3

1 Conceptually with 2's complement encoding, the least significant byte pattern of int 255 or 0000...0000_1111_11112 is preserved as signed char where the lead bit is a sign (or -128 place).

Upvotes: 6

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