Challenge #21 description it's quite simple
This is easy. What does the following code do?
We can assume this will be easy.
Analysis
Here's the assembly code we need to analyze
f1:
push r12
push rbp
mov rbp, rdi
push rbx
mov rbx, rsi
;; strlen(rdi);
call strlen
;; strlen(rbx);
mov rdi, rbx
mov r12, rax
call strlen
;; size_1 -= size_2;
sub r12, rax
mov rsi, rbx
lea rdi, [rbp+0+r12]
call strcmp
pop rbx
test eax, eax
pop rbp
sete al
pop r12
ret
)
We have two calls to strlen and one to strcmp. With this in mind we can assume we receive two strings as arguments of our function.
Right after the calls to strlen
we have the following instruction
sub r12, rax
In r12
we stored the result of our first call to strlen
, so with this instruction we are computing the difference in length of these two strings.
The following three instructions perform a comparison in size of strings rbx
and [rbp+0+r12]
.
mov rsi, rbx
lea rdi, [rbp+0+r12]
call strcmp
In r12
we have the result of the difference, so if for str1
we have str1_size
, and for str2
we have str2_size
, then in r12
we have str1_size - str2_size
. Hence these three instructions can be expressed in this way
size_t s1 = strlen(str1);
size_t s2 = strlen(str2);
strcmp((str1+(s1-s2)), str2);
Our last step it's quite simple also, in case our comparison with strcmp
was 0 we return 1, otherwise we return 0.
Formal description
The function takes two strings as arguments and return 1 if one string it's the suffix of the other, and 0 otherwise.
Conclusion
Coming from the frustration with challenge #19, making this one was quite easy.
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