this program meant to add 2 float nums and display their respective results but program gives unwanted results and i dont understand why is it necessary to allocate 16 bytes for a float rather as a double occupies 8 bytes so why not allocate only 8 bytes for float?
.text
.globl main
.type main, @function
main:
subl $12, %esp # allocate enough memory for a floating point value
flds (V0) # load loading single precision variable 1
flds (V1) # load single precision variable 2
fadd %st(1), %st(0) # add both of them [ NOTE: reg ST(0) contains correct result ]
fstpl (%esp) # store the float
pushl $.LC0
call printf
addl $16, %esp # return allocated mem to OS [ 12 + 4 = 16 ]
ret
.LC0: .string "%f\n"
V0: .float 9.3
V1: .float 9.4
Looks buggy to me; ESP is 16-byte aligned before the call to main (which pushes a 4-byte return address), so it should be using sub $12, %esp to recreate 16-byte alignment as required by the i386 System V ABI (assuming you're on Linux) before the call to printf.
That's what you'd see if you compiled a C function with a C compiler.
I'd also recommend using flds for the single-precision loads and fstpl for the double-precision store, to make the sizes explicit. This is also buggy; the default is the same for load and store, and this program needs float load and double store. (The default is single-precision float, dword)
(printf "%f" takes a double because there's no way of passing a float in C to a variadic function: the default promotions apply.)