CODEWITHSUNDEEP
assemblyx86att
pneuc4353
pneuc4353

Reputation: 21

How do I find the end of a string in x86 assembly?

The code below is an i86 assembly function that will be called in c, count(char *string, char c). I'm storing the string in %esi and the character in %ecx, I'm not sure how to tell the program to stop looping when the string is over(when bytes are 0) because the cmpl function doesn't allow me to do that (operand type mismatch). I'm not sure if I'm even storing the string correctly or iterating through the string correctly. Any help would be greatly appreciated



.globl count
.text
#first argument is string, second is char
count:  
    push %ebp       #push old ebp to top of stack
    movl %esp, %ebp     #point ebp to same thing as esp, both pointing to old ebp

    #saving live registers
    push %esi
    push %edi
    push %ebx   

    #take values from stack
    movl $0, %eax       #set count(eax) to 0
    movl 12(%ebp), %ecx     #move letter to be counted to ecx
    movl 8(%ebp), %esi  #move string to esi

lettercheck:
    cmpl %esi, %ecx #compare current letter to ecx(letter to be counted)
    incl %esi       #next letter
    je inccount     #if match inccount 
    cmpl %esi, $0       #if no match and end of string
    je end
    jmp lettercheck     #if not end of string and no match  

inccount:
    incl eax            #inc count of char
    cmpl %esi, $0       #check for end of string
    jne lettercheck     #if match and not EOS lettercheck
    jmp end         #else jmp end
end:
    pop %ebx
    pop %edi
    pop %esi
    movl %ebp, %esp
    pop %ebp
    ret

Upvotes: 1

Views: 2029

Answers (1)

Chris Dodd
Chris Dodd

Reputation: 126195

%esi holds a pointer, so you want to compare the byte it points at to the bottom byte of %ecx, not compare the pointer. So your comparison instruction should be

cmpb (%esi), %cl

The b mean byte instead of longword, and the parens mean the value in memory pointed at by the register, rather than the register. To check for the end of the string, you want

cmpb $0, (%esi)

for all the same reasons, plus the fact that an immediate can only be the first operand to a comparison.

Upvotes: 1

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