将movsbl汇编为C代码

Question

I'm trying to convert the following optimised assembly code to C code but keeping as close to the assembly code as possible. I have no idea how to change the movsbl call to C code. My understanding is that it moves a byte with a zero extension into a 32bit register. I have included comments as to what I believe to be happening in the assembly code.

file    "my_sieve.c"
.text
    .p2align 4,,15
    .globl  my_sieve
    .type   my_sieve, @function
my_sieve:
    pushl   %ebp
    movl    $4, %eax            #eax = 4
    pushl   %edi
    movl    $1, %ebp            #ebp = 1
    pushl   %esi
    movl    $2, %esi            #esi = 2
    pushl   %ebx
    movl    24(%esp), %edi      #edi = max
    cmpl    $3, %edi            #if edi > 3
    jg  .L9                     #go to L9
    jmp .L1                     #otherwise go to L1
    .p2align 4,,7
    .p2align 3
.L6:
    addl    $1, %esi            #esi + 1
    movl    %esi, %eax          #eax = esi
    imull   %esi, %eax          #eax*esi (i*i)
    cmpl    %edi, %eax          #if i>max
    jg  .L1                     #go to L1
.L9:
    movl    20(%esp), %ebx      #ebx = composite
    movl    %esi, %edx          #edx = esi
    sarl    $3, %edx            #shift edx right 3 bits
    movsbl  (%ebx,%edx), %ecx   #??
    movl    %esi, %edx          #edx = esi
    andl    $7, %edx            #bit-wise AND between 7 & edx, stores result in edx
    btl %edx, %ecx              #copies edx to ecx  
    jc  .L6                     #jump to L6 if carry
    .p2align 4,,7
    .p2align 3
.L7:
    movl    %eax, %ecx          #ecx = eax
    movl    %ebp, %ebx          #ebx = ebp
    andl    $7, %ecx            #bit-wise AND between 7 & ecx, stores result in ecx
    movl    %eax, %edx          #edx = eax
    sall    %cl, %ebx           #left shift ebx the number of bits held in cl 
    addl    %esi, %eax          #eax + esi
    movl    %ebx, %ecx          #ecx = ebx
    movl    20(%esp), %ebx      #ebx = composite
    sarl    $3, %edx            #sign preserving right shift edx 3 bits
    orb %cl, (%ebx,%edx)        #8-bit logical OR 
    cmpl    %eax, %edi          #compare edi & eax
    jge .L7                     #jump if greater or equal
    jmp .L6                     #otherwise jump to L6
    .p2align 4,,7
    .p2align 3
.L1:
    popl    %ebx
    popl    %esi
    popl    %edi
    popl    %ebp
    ret                         #return
    .size   my_sieve, .-my_sieve
    .ident  "GCC: (Ubuntu 4.8.2-19ubuntu1) 4.8.2"
    .section    .note.GNU-stack,"",@progbits

And my incomplete attempt at creating C code. I need help to fill in the blanks or someone to tell me what I'm attempting is completely wrong.

void my_sieve(char *composite, int max){
long ebp, eax, edi, esi, edi, ecx, edx;


eax = 4;
ebp = 1;
esi = 2;
edi = max;

if(edi > 3)
goto L9;
else
goto L1;

L6:
    esi += 1;
    eax = esi;
    eax = eax*esi;
    if(eax > edi)
    goto L1;

L9:
    ebx = composite;
    edx = esi;
    //right shift
    edx = edx >> 3;

    //movsbl

    edx = esi;
    //bit-wise
    edx = edx & 7;

    edx = ecx;
    if(carry)
    goto L6;

L7:
    ecx = eax;
    ebx = ebp;
    //bit-wise
    ecs = ecx & 7;

    edx = eax;
    //left shift
    ebx = ebx & 0xFF;

    eax += esi;
    ecx = ebx;
    ebx = composite;
    //right shift
    edx = edx >> 3;

    //8 bit logical OR

    if(edi >= eax)
    goto L7;
    else
    goto L6;

L1:
    return;
}

Answer 1

I think (int)((int8_t)value) would do it. Casting between unsigned and signed types of the same size has no effect on the bit pattern. Casting a smaller signed type to a larger one causes a sign extension.