intel-syntax

I am quite new regarding the assembly and I am trying to work with a program. So whenever I try to compile it, I get the error for the line, as listed under the comments in the code. I am wondering if anyone has any ideas why NASM detects this errors when I am defining some things for the rest of the assembly code? Maybe it has to do something with how the main is defined? P.S. I listed just the first part of the code, since the program is quite long. Thank you for the help .xlist ;attempt to define a local label before any non-local labels include stdlib.a ; parser: instruction expected

When reviewing gdb output and looking at the assembly calls, usually I can find a command using hard-coded values to determine whether the registers are being loaded right to left or vice versa. Usually something like the following: sub rsp, 16 or sub 16, rsp But other times, no values like above are visible. All I see are calls like the following : (gdb) disassemble Dump of assembler code for function main: 0x0000000100000f54 <main+4>: mov $rdi,%r15 0x0000000100000f59 <main+9>: mov $rsi,%r14 0x0000000100000f60 <main+16>: mov $rdx,%r13 0x0000000100000f67 <main+23>: mov $ecx,$r12d End of

The general form for memory addressing in Intel and AT&T Syntax is the following: [base + index*scale + disp] disp(base, index, scale) My questions are the following: Can base and index be any register? What values can scale take, is it 1, 2, 4 and 8 (with 1 being the default)? Are index and disp interchangeable (with the only difference being that index is a register while disp is an immediate value)? Michael This is described in Intel's manual: 3.7.5 Specifying an Offset The offset part of a memory address can be specified directly as a static value (called a displacement) or through an

Can anyone explain to me why every constant in AT&T syntax has a '$' in front of it? Why do all registers have a '%'? Is this just another attempt to get me to do a lot of lame typing? Also, am I the only one that finds: 16(%esp) really counterintuitive compared to [esp+16] ? I know it compiles to the same thing but why would anyone want to type a lot of '$' and '%'s without a need to? - Why did GNU choose this syntax as the default? Another thing, why is every instruction in at&t syntax preceded by an: l? - I do know its for the operand sizes, however why not just let the assembler figure

Consider the following variable reference in x64 Intel assembly, where the variable a is declared in the .data section: mov eax, dword ptr [rip + _a] I have trouble understanding how this variable reference works. Since a is a symbol corresponding to the runtime address of the variable (with relocation), how can [rip + _a] dereference the correct memory location of a ? Indeed, rip holds the address of the current instruction, which is a large positive integer, so the addition results in an incorrect address of a ? Conversely, if I use x86 syntax (which is very intuitive): mov eax, dword ptr [

Why doesn't this code set temp to 1? How do I actually do that? int temp; __asm__( ".intel_syntax;" "mov %0, eax;" "mov eax, %1;" ".att_syntax;" : : "r"(1), "r"(temp) : "eax"); printf("%d\n", temp); You want temp to be an output, not an input, I think. Try: __asm__( ".intel_syntax;" "mov eax, %1;" "mov %0, eax;" ".att_syntax;" : "=r"(temp) : "r"(1) : "eax"); This code does what you are trying to achieve. I hope this helps you: #include <stdio.h> int main(void) { /* Compile with C99 */ int temp=0; asm ( ".intel_syntax;" "mov %0, 1;" ".att_syntax;" : "=r"(temp) : /* no input*/ ); printf("temp=%d