C++ double 浮点精度问题?
起因
早上起来玩CF,发现竟然是一个浮点数问题! 真的好害怕.
题目条件说输入的参数
c
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m
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p
=
1
c+m+p=1
c+m+p=1,然后我就在代码里加上了assert(c+m+p==1.),结果就是Runtime error.
经过
反复测试,反复修改,我将代码精简如下.
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22/*D:codecppexamplefloat_precision>type main.c*/ #include <assert.h> typedef unsigned char bool; bool directCompare(double a,double b,double c,double r) { return a+b+c==r; } bool answerCompare(double a,double b,double c,double r) { double ans=a+b+c; return ans==r; } int main() { assert( directCompare(0.4,0.2,0.4,1.) == answerCompare(0.4,0.2,0.4,1.) ); return 0; }
然后,编译运行.
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5D:codecppexamplefloat_precision>gcc main.c -o a D:codecppexamplefloat_precision>a Assertion failed: directCompare(0.4,0.2,0.4,1.) == answerCompare(0.4,0.2,0.4,1.), file main.c, line 18
神奇的一幕发生了.
结果竟然不一样.
于是我只得
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2D:codecppexamplefloat_precision>gcc main.c -S
看汇编吧,希望我能看懂.
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136.file "main.c" .text .globl _directCompare .def _directCompare; .scl 2; .type 32; .endef _directCompare: pushl %ebp movl %esp, %ebp subl $32, %esp movl 8(%ebp), %eax movl %eax, -8(%ebp) movl 12(%ebp), %eax movl %eax, -4(%ebp) movl 16(%ebp), %eax movl %eax, -16(%ebp) movl 20(%ebp), %eax movl %eax, -12(%ebp) movl 24(%ebp), %eax movl %eax, -24(%ebp) movl 28(%ebp), %eax movl %eax, -20(%ebp) movl 32(%ebp), %eax movl %eax, -32(%ebp) movl 36(%ebp), %eax movl %eax, -28(%ebp) fldl -8(%ebp) faddl -16(%ebp) faddl -24(%ebp) fldl -32(%ebp) fucomip %st(1), %st setnp %al movl $0, %edx fldl -32(%ebp) fucomip %st(1), %st fstp %st(0) cmovne %edx, %eax leave ret .globl _answerCompare .def _answerCompare; .scl 2; .type 32; .endef _answerCompare: pushl %ebp movl %esp, %ebp subl $48, %esp movl 8(%ebp), %eax movl %eax, -24(%ebp) movl 12(%ebp), %eax movl %eax, -20(%ebp) movl 16(%ebp), %eax movl %eax, -32(%ebp) movl 20(%ebp), %eax movl %eax, -28(%ebp) movl 24(%ebp), %eax movl %eax, -40(%ebp) movl 28(%ebp), %eax movl %eax, -36(%ebp) movl 32(%ebp), %eax movl %eax, -48(%ebp) movl 36(%ebp), %eax movl %eax, -44(%ebp) fldl -24(%ebp) faddl -32(%ebp) faddl -40(%ebp) fstpl -8(%ebp) fldl -8(%ebp) fldl -48(%ebp) fucomip %st(1), %st fstp %st(0) setnp %al movl $0, %edx fldl -8(%ebp) fldl -48(%ebp) fucomip %st(1), %st fstp %st(0) cmovne %edx, %eax leave ret .def ___main; .scl 2; .type 32; .endef .section .rdata,"dr" LC4: .ascii "main.c" .align 4 LC5: .ascii "directCompare(0.4,0.2,0.4,1.) == answerCompare(0.4,0.2,0.4,1.)" .text .globl _main .def _main; .scl 2; .type 32; .endef _main: pushl %ebp movl %esp, %ebp pushl %ebx andl $-16, %esp subl $32, %esp call ___main fld1 fstpl 24(%esp) fldl LC2 fstpl 16(%esp) fldl LC3 fstpl 8(%esp) fldl LC2 fstpl (%esp) call _directCompare movl %eax, %ebx fld1 fstpl 24(%esp) fldl LC2 fstpl 16(%esp) fldl LC3 fstpl 8(%esp) fldl LC2 fstpl (%esp) call _answerCompare cmpb %al, %bl je L6 movl $18, 8(%esp) movl $LC4, 4(%esp) movl $LC5, (%esp) call __assert L6: movl $0, %eax movl -4(%ebp), %ebx leave ret .section .rdata,"dr" .align 8 LC2: .long -1717986918 .long 1071225241 .align 8 LC3: .long -1717986918 .long 1070176665 .ident "GCC: (tdm-1) 5.1.0" .def __assert; .scl 2; .type 32; .endef
终于看懂了,浮点数加减法使用的是x87 FPU指令.而该计算环境使用的80位的浮点精度,在数据传输时将double存储的64位自动转化为80位,然后计算完再转换回来.
directCompare在FPU上进行比较,而answerCompare比较之前将数据转换为64位,然后又转换为80位,产生了精度损失(?结果却算对了).
结果
看了
-
intel的文档:
- Intel® 64 and IA-32 architectures software developer’s manual volume 1: Basic architecture
- Intel® 64 and IA-32 architectures software developer’s manual combined volumes 2A, 2B, 2C, and 2D: Instruction set reference, A- Z
-
gdb的文档:
- GDB User Manual
-
了解了:
- gdb命令:
- ni: 下一条指令
- si: 进入call
- set confirm off: k,q不提示
- set disassemble-next-line
- x/nfu: 检查内存
- print: 显示
- $ps: program status register
- x86指令:
- setnp: set if not PF
- cmovne: move if not equal
- X87 FPU 指令:
- fld1: 加载1到%st
- fld: 加载数据到%st
- fadd: 加数据到%st
- fucomip: unordered 比较%st(0) %st(i) 弹栈
- fstp: 将%st的数据装入寄存器中 弹栈
- gdb命令:
最后
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