# x86 Disassembly/Floating Point Examples

## Example: Floating Point Arithmetic edit

Here is the C source code, and the GCC assembly listing of a simple C language function that performs simple floating-point arithmetic. Can you determine what the numerical values of LC5 and LC6 are?

```
__fastcall double MyFunction2(double x, double y, float z)
{
return (x + 1.0) * (y + 2.0) * (z + 3.0);
}
```

```
.align 8
LC5:
.long 0
.long 1073741824
.align 8
LC6:
.long 0
.long 1074266112
.globl @MyFunction2@20
.def @MyFunction2@20; .scl 2; .type 32; .endef
@MyFunction2@20:
pushl %ebp
movl %esp, %ebp
subl $16, %esp
fldl 8(%ebp)
fstpl -8(%ebp)
fldl 16(%ebp)
fstpl -16(%ebp)
fldl -8(%ebp)
fld1
faddp %st, %st(1)
fldl -16(%ebp)
fldl LC5
faddp %st, %st(1)
fmulp %st, %st(1)
flds 24(%ebp)
fldl LC6
faddp %st, %st(1)
fmulp %st, %st(1)
leave
ret $20
```

For this, we don't even need a floating-point number calculator, although you are free to use one if you wish (and if you can find a good one). LC5 is added to [ebp - 16], which we know to be y, and LC6 is added to [ebp - 24], which we know to be z. Therefore, LC5 is the number "2.0", and LC6 is the number "3.0". Notice that the **fld1** instruction automatically loads the top of the floating-point stack with the constant value "1.0".