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📄 sum.yo

📁 深入理解计算机系统 的lab
💻 YO
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  0x000:              | 	.pos 0  0x000: 308400010000 | init:	irmovl Stack, %esp  0x006: 308500010000 | 	irmovl Stack, %ebp  0x00c: 702c000000   | 	jmp Main                      |                       | # Sample linked list  0x014:              | .align 4  0x014:              | ele1:  0x014: 0a000000     |          .long 0x00a  0x018: 1c000000     |          .long ele2  0x01c:              | ele2:  0x01c: b0000000     |          .long 0x0b0  0x020: 24000000     |          .long ele3  0x024:              | ele3:  0x024: 000c0000     |          .long 0xc00  0x028: 00000000     |          .long 0                      |   0x02c: 308004000000 | Main:	irmovl $4, %eax  0x032: a008         | 	pushl %eax  0x034: 308214000000 | 	irmovl ele1, %edx  0x03a: a028         | 	pushl %edx  0x03c: 8042000000   | 	call sum_list  0x041: 10           | 	halt                      |   0x042:              | sum_list:	  0x042: a058         | 	pushl %ebp  0x044: 2045         | 	rrmovl %esp, %ebp  0x046: 501508000000 | 	mrmovl 8(%ebp), %ecx	#ecx=ls  0x04c: 6300         | 	xorl %eax, %eax		#val=0  0x04e:              | Loop:	  0x04e: 6211         | 	andl %ecx, %ecx  0x050: 7368000000   | 	je End  0x055: 506100000000 | 	mrmovl (%ecx), %esi	#get* ls  0x05b: 6060         | 	addl %esi, %eax		#add to val                      | #	irmovl $4, %ebx		                      | #	addl %ebx, %ecx		#ls--  0x05d: 501104000000 | 	mrmovl 4(%ecx), %ecx	#ls=ls->next  0x063: 704e000000   | 	jmp Loop  0x068:              | End:  0x068: b058         | 	popl %ebp  0x06a: 90           | 	ret  0x100:              | 	.pos 0x100  0x100:              | Stack:                      | 

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