Question

Foo1:

pushl %ebp
movl %esp,%ebp
movl 8(%ebp),%eax
sall $4,%eax
subl 8(%ebp),%eax
movl %ebp,%esp
popl %ebp
ret
foo1 corresponds to which C code choice?
Group of answer choices
int choice(int x){return (x >> 31); }
int choice(int x) {return x / 16; }
int choice(int x){return (x + 15) /4 }
int choice(int x){return (x << 31) & 1; }
int choice(int x){return 15 * x;}

Answer 1

The assembly code corresponds to the C code int choice(int x) {return (x + 15) /4; }. It takes an input parameter x, adds 15 to it, and then divides the result by 4.

Step-by-step explanation:

The given assembly code corresponds to the following C code:

int choice(int x) {return (x + 15) /4; }

The assembly code performs the following steps:

1. Pushes the value of %ebp onto the stack.
2. Moves the value of %esp into %ebp.
3. Moves the value at the memory address 8(%ebp) into %eax.
4. Left shifts the value in %eax by 4 bits.
5. Subtracts the value at the memory address 8(%ebp) from the value in %eax.
6. Moves the value of %ebp into %esp.
7. Pops the value of %ebp from the stack.
8. Returns the value in %eax.

The C code equivalent of the assembly code is int choice(int x) {return (x + 15) /4; }. This C code takes an input parameter x, adds 15 to it, and then divides the result by 4.