assembly

问题 Is there a way to perform a bitwise NAND operation on the bits in two registers in ARM7, either with the existing AND, OR and EOR operations or other instructions? 回答1: Sure; AND the two registers and then EOR the result with all 1's (for the negation). 回答2: and then mvn (move not). From GCC explorer int nand(int a, int b) { return ~(a & b); } nand(int, int): and r0, r0, r1 mvn r0, r0 bx lr 来源： https://stackoverflow.com/questions/21207561/nand-logical-bitwise-operation-in-arm

问题 I'm getting a undefined reference to _printf when building an assembly program that defines its own _start instead of main , using NASM on x86-64 Ubuntu Build commands: nasm -f elf64 hello.asm ld -s -o hello hello.o hello.o: In function `_start': hello.asm:(.text+0x1a): undefined reference to `_printf' MakeFile:4: recipe for target 'compile' failed make: *** [compile] Error 1 nasm source: extern _printf section .text global _start _start: mov rdi, format ; argument #1 mov rsi, message ;

问题 I'm getting a undefined reference to _printf when building an assembly program that defines its own _start instead of main , using NASM on x86-64 Ubuntu Build commands: nasm -f elf64 hello.asm ld -s -o hello hello.o hello.o: In function `_start': hello.asm:(.text+0x1a): undefined reference to `_printf' MakeFile:4: recipe for target 'compile' failed make: *** [compile] Error 1 nasm source: extern _printf section .text global _start _start: mov rdi, format ; argument #1 mov rsi, message ;

问题 Why is it that 32-bit C pushes all function arguments straight onto the stack while 64-bit C puts the first 6 arguments into registers and the rest on the stack? So the 32-bit stack would look like: ... arg2 arg1 return address old %rbp While the 64-bit stack would look like: ... arg8 arg7 return address old %rbp arg6 arg5 arg4 arg3 arg2 arg1 So why does 64-bit C do this? Isn't it much easier to just push everything to the stack instead of put the first 6 arguments in registers just to move

问题 Brand new to assembly need some help with unsigned arithmetic. Converting from a C program is that means anything. Using: Linux NASM x86 (32 Bit) I want to read in a number from the user. I want this number to be unsigned. When I enter a number above the signed integer limit and use info registers, I notice that my register storing that is negative which means an overflow happened. (Obviously number entered is below max unsigned int) How do I treat this register as unsigned so I can do

问题 Brand new to assembly need some help with unsigned arithmetic. Converting from a C program is that means anything. Using: Linux NASM x86 (32 Bit) I want to read in a number from the user. I want this number to be unsigned. When I enter a number above the signed integer limit and use info registers, I notice that my register storing that is negative which means an overflow happened. (Obviously number entered is below max unsigned int) How do I treat this register as unsigned so I can do

问题 Jump's based on comparing signed integers use the Zero, Sign, and Overflow flag to determine the relationship between operands. After CMP with two signed operands, there are three possible scenario's: ZF = 1 - Destination = Source SF = OF - Destination > Source SF != OF - Destination < Source I'm having trouble understanding scenario 2 and 3. I've worked through the possible combinations and see that they do work - but I still can't figure out why they work. Can anyone explain why a

问题 Jump's based on comparing signed integers use the Zero, Sign, and Overflow flag to determine the relationship between operands. After CMP with two signed operands, there are three possible scenario's: ZF = 1 - Destination = Source SF = OF - Destination > Source SF != OF - Destination < Source I'm having trouble understanding scenario 2 and 3. I've worked through the possible combinations and see that they do work - but I still can't figure out why they work. Can anyone explain why a

问题 Jump's based on comparing signed integers use the Zero, Sign, and Overflow flag to determine the relationship between operands. After CMP with two signed operands, there are three possible scenario's: ZF = 1 - Destination = Source SF = OF - Destination > Source SF != OF - Destination < Source I'm having trouble understanding scenario 2 and 3. I've worked through the possible combinations and see that they do work - but I still can't figure out why they work. Can anyone explain why a

问题 Jump's based on comparing signed integers use the Zero, Sign, and Overflow flag to determine the relationship between operands. After CMP with two signed operands, there are three possible scenario's: ZF = 1 - Destination = Source SF = OF - Destination > Source SF != OF - Destination < Source I'm having trouble understanding scenario 2 and 3. I've worked through the possible combinations and see that they do work - but I still can't figure out why they work. Can anyone explain why a