MIPS translator in C++, how to convert to hex
I am creating a translator in C++ that accepts MIPS assembly instructions such as "add 1, 2, 3" as in add register 2 and 3 and place the result in register 1. Eventually, I save everything into a string representing the machine code that is supposed to be the output, however I want to display it in hexadecimal format, instead it outputs in decimal
assembly = "000000" + m[3].str() + m[4].str() + m[2].str() + "00000" + funct;
cout << assembly << endl;
The literal output is correct = 00000023100000100000
However I want it to output as = 00430820
Can anyone please help me? Thank you!
you mix binary, with integers
the encoding for add s,t,d in BINARY is
"0000 00ss ssst tttt dddd d000 0010 0000"
so you'd have to concat "000000" + binary(2) + binary(3) + binary(1) + "00000100000"
where binary(x) is the binary represenation with fix 5 digits
in your case
000000 + 00010 + 00011 + 00001 + 00000100000
= 00000000010000110000100000100000, which converts to
00430820h ("0000"=0, "0000"=0, "0100"=4, "0011" = 3, "0000"=0, "1000"=8, "0010"=2, "0000"=0)
if you take a closer look at the encoding, you see
"000000ssssstttttddddd00000100000"
consists of 4 parts. first is constant which defines the instruction, and 3 are defining the registers to be used
"00000000000000000000000000100000" = 0x00000020
+ "000000sssss000000000000000000000" (which is s left shifted by 21)
+ "00000000000ttttt0000000000000000" (which is t left shifted by 16)
+ "0000000000000000ddddd00000000000" (which is d left shifted by 11)
so the easiest way to encode this instruction is:
uint32_t i = 0x00000020 + (s<<21) + (t<<16) + (d<<11);
print this in hex, and you get: 00430820