How to remove a char from a 4-byte string in MIPS32 assembly using a shift?

Question

I'm new to MIPS32 assembly and trying to delete a character in a string (delete the first character, specifically) stored in the.data section but have no clue how to do so.

In the following line of code, is there a way to make it so that test just equals "bc" instead of "abc"

test:           .asciiz     "abc"

Is this simply a matter of using something like logical shifting left by 2 to remove the first char, or do I need to offset by something, or is there an opcode to just straight delete it?

Answer 1

As Need to remove all non letter elements from a string in assembly (for x86) explains, removing a character in a string means copying over the whole rest of the string.

In your case it's just 3 bytes left out of the 4 (including the terminating 0 ). So yes, you could do this just by shifting the word by 8 bits (1 byte). Especially if you make sure test is word-aligned with .p2align 2 before it, so you can safely lw and sw all 4 bytes with one load.

For little-endian MIPS (like MARS simulates), that would be a right shift because the first byte in memory is the least significant. And right shifts shift out the low (least significant) bits.

For big-endian MIPS (most significant byte first, like some real MIPS CPUs operate), that would be a left shift, removing the most significant byte and shifting the low bits up.

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Note that this will leave the word at test being 'b', 'c', 0, 0 . So yes, as an implicit-length string it's "bc" .

Also note that if you just had a pointer in a register, you could get a pointer to "bc" by simply incrementing it by 1 instead of modifying memory. Like addiu $t0, $t0, 1 .

Or equivalently, la $t0, test+1 is a pointer to 1 byte past the start.