MARS MIPS multiplication of two single-precision IEEE 754 standard floating-point numbers

Question

Write a program to perform the multiplication of two single-precision IEEE 754 standard floating-point numbers without using MIPS floating-point arithmetic instructions. The input data is read from a file stored in binary format on the disk named FLOAT2.BIN (2 values x 4 bytes = 8 bytes).

.data
    num1:   .float 2.5   
    num2:   .float 3.75    

.text
.globl main
main:
 
    lwc1 $f0, num1

    lwc1 $f1, num2

    li $t0, 0x80000000      
    mtc1 $t0, $f2          

    lui $t0, 0x3f80         
    mtc1 $t0, $f3         

    sll $t0, $t0, 1      
    mtc1 $t0, $f4        

    mul.s $f5, $f0, $f1    

    mul.s $f6, $f2, $f5    
    mul.s $f7, $f3, $f5     
    mul.s $f8, $f4, $f5    

    add.s $f9, $f6, $f7    
    add.s $f10, $f9, $f8   
    mov.s $f12, $f10       

    li $v0, 2             
    mov.s $f12, $f10      
    syscall                


    li $v0, 10             
    syscall                

I don't know why the result is infinity.

Answer 1

I don't know why the result is infinity.

The value in $f4 is 0x7f000000, which is 1.7014118346e+38, notably, a very large number, whose exponent is at the limit of single precision floating point.

The value in $f5 is 0x41160000, which is 9.375.

The program then multiplies these two values together using

mul.s $f8, $f4, $f5   

Numerically the result is 1.5950735949375e+39, however, this value overflows the single precision floating point format, so is converted to "infinity", which is stored in $f8.

Later $f10 is generated by adding $f9 to $f8,

add.s $f10, $f9, $f8   

This propagates the infinity/overflow from $f8 into $f10, and that's what is being printed.

You can observe those values in the floating point registers using a good floating point calculator like: https://www.h-schmidt.net/FloatConverter/IEEE754.html