Question

We found some strange values being produced, a small test case is below.
This prints “FFFFFFFFF9A64C2A” . Meaning the unsigned long long seems to have been sign extended.
But why ?
All the types below are unsigned, so what’s doing the sign extension ? The expected output
would be “F9A64C2A”.

#include <stdio.h>

int main(int argc,char *argv[])
{
    unsigned char a[] = {42,76,166,249};

    unsigned long long ts;
    ts = a[0] | a[1] << 8U | a[2] << 16U | a[3] << 24U;

    printf("%llX\n",ts);


    return 0;

}

Answer 1

In the expression a[3] << 24U, the a[1] has type unsigned char. Now, the “integer promotion” converts it to int because:

The following may be used in an expression wherever an int or unsigned int may
be used:

[…]

If an int can represent all values of the original type, the value is converted to
an int;
otherwise, it is converted to an unsigned int.

((draft) ISO/IEC 9899:1999, 6.3.1.1 2)

Please note also that the shift operators (other than most other operators) do not do the “usual arithmetic conversions” converting both operands to a common type. But

The type of the result is that of the promoted left operand.

(6.5.7 3)

On a 32 bit platform, 249 << 24 = 4177526784 interpreted as an int has its sign bit set.

Just changing to

ts = a[0] | a[1] << 8 | a[2] << 16 | (unsigned)a[3] << 24;

fixes the issue (The suffix Ufor the constants has no impact).