Why doesn't C have binary literals?

According to Rationale for International Standard - Programming Languages C §6.4.4.1 Integer constants

A proposal to add binary constants was rejected due to lack of precedent and insufficient utility.

It's not in standard C, but GCC supports it as an extension, prefixed by 0b or 0B:

 i = 0b101010;

See here for detail.

This is what pushed hexadecimal to be... hexadecimal. The "... primary use of hexadecimal notation is a human-friendly representation of binary-coded values in computing and digital electronics ...". It would be as follows:

val1 |= 0xF;val2 &= 0x40;val3 |= ~0x10;

Hexadecimal:

1. One hex digit can represent a nibble (4 bits or half an octal).
2. Two hex digits can represent a byte (8 bits).
3. Hex is much more compact when scaling to larger masks.

With some practice, converting between hexadecimal and binary will become much more natural. Try writing out your conversions by hand and not using an online bin/hex notation converter -- then in a couple days it will become natural (and quicker as a result).

Aside: Even though binary literals are not a C standard, if you compile with GCC it is possible to use binary literals, they should be prefixed with '0b' or '0B'. See the official documentation here for further information. Example:

int b1 = 0b1001; // => 9int b2 = 0B1001; // => 9

All of your examples can be written even more clearly:

val1 &= (1 << 4) - 1; //clear high nibbleval2 |= (1 << 6); //set bit 6val3 &=~(1 << 3); //clear bit 3

(I have taken the liberty of fixing the comments to count from zero, like Nature intended.)

Your compiler will fold these constants, so there is no performance penalty to writing them this way. And these are easier to read than the 0b... versions.