CODEWITHSUNDEEP
c++c
user318747
user318747

Reputation: 1428

Why does this hex value get output as a negative number?

char buffer_b[5] = { 0xDA, 0x00, 0x04, 0x00, 0x07 };
printf("%d\n%d\n%d", buffer_b[0], buffer_b[2], buffer_b[4]);

This gives me output:

-38
4
7

However I am expecting:

218
4
7

Thanks.

Upvotes: 2

Views: 534

Answers (3)

Norman Ramsey
Norman Ramsey

Reputation: 202495

When the char 0xDA is promoted to int to pass to printf, the compiler is doing a sign-extension, converting it to 0xffffffda, which is the 32-bit representation of -38. You were expecting it to be zero-extended to 0x000000da. To control how the compiler extends a character, you have declare it as signed char or unsigned char. Signed integer types are widened by sign-extending, and unsigned integer types are widened by zero-extending.

You can't predict how any particular compiler will treat an unqualified char, or if it will be the same in the next release of the compiler.

Upvotes: 2

Rob Kennedy
Rob Kennedy

Reputation: 163257

Evidently, char is signed in your environment. (That's a detail that can vary from one implementation to the next, and some compilers even offer you an option through a command-line switch.) The number you're printing is 0xDA, which has the most significant bit set, so its value is negative. When the compiler passes that value to printf, it promotes the (signed) char value to type int, and it retains its negativity. You used the %d format string, which tells printf to interpret its argument as a signed value.

To treat the value as unsigned, you should at a minimum use the %u format string. Then either change your array's element type to be an unsigned type, such as unsigned char or uint8_t, or type-cast the printf argument to unsigned.

Upvotes: 8

Edward Strange
Edward Strange

Reputation: 40859

char is signed. Use unsigned char.

use %ud also.

Upvotes: 8

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