x86 Disassembly/Floating Point Examples

x86 Disassembly

Example: Floating Point Arithmetic

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".