What does %call16 and .reloc mean in mips? - c

Question

I have converted a c code into mips32 using gcc compiler but there are some parts that I did not understand.
This is c code:
int main()
{
float fibSquared;
int F[10] = {0};
F[0] = 0;
F[1] = 1;
for(int i = 2; i < 10 ; i++)
F[i] = F[i-1] + F[i-2];
fibSquared = sqrt(F[9]);
printf("%f",fibSquared);
return 0;
}
This is mips code:
.file 1 ""
.section .mdebug.abi32
.previous
.nan legacy
.module fp=32
.module nooddspreg
.abicalls
.rdata
.align 2
$LC0:
.ascii "%f\000"
.text
.align 2
.globl main
.set nomips16
.set nomicromips
.ent main
.type main, #function
main:
.frame $fp,80,$31 # vars= 48, regs= 2/0, args= 16, gp= 8
.mask 0xc0000000,-4
.fmask 0x00000000,0
.set noreorder
.cpload $25
.set nomacro
addiu $sp,$sp,-80
sw $31,76($sp)
sw $fp,72($sp)
move $fp,$sp
.cprestore 16
movz $31,$31,$0
sw $0,32($fp)
sw $0,36($fp)
sw $0,40($fp)
sw $0,44($fp)
sw $0,48($fp)
sw $0,52($fp)
sw $0,56($fp)
sw $0,60($fp)
sw $0,64($fp)
sw $0,68($fp)
sw $0,32($fp)
li $2,1 # 0x1
sw $2,36($fp)
li $2,2 # 0x2
sw $2,24($fp)
b $L2
nop
$L3:
lw $2,24($fp)
nop
addiu $2,$2,-1
sll $2,$2,2
addiu $3,$fp,24
addu $2,$3,$2
lw $3,8($2)
lw $2,24($fp)
nop
addiu $2,$2,-2
sll $2,$2,2
addiu $4,$fp,24
addu $2,$4,$2
lw $2,8($2)
nop
addu $3,$3,$2
lw $2,24($fp)
nop
sll $2,$2,2
addiu $4,$fp,24
addu $2,$4,$2
sw $3,8($2)
lw $2,24($fp)
nop
addiu $2,$2,1
sw $2,24($fp)
$L2:
lw $2,24($fp)
nop
slt $2,$2,10
bne $2,$0,$L3
nop
lw $2,68($fp)
nop
mtc1 $2,$f0
nop
cvt.d.w $f0,$f0
mov.d $f12,$f0
lw $2,%call16(sqrt)($28)
nop
move $25,$2
.reloc 1f,R_MIPS_JALR,sqrt
1: jalr $25
nop
lw $28,16($fp)
cvt.s.d $f0,$f0
swc1 $f0,28($fp)
lwc1 $f0,28($fp)
nop
cvt.d.s $f0,$f0
mfc1 $7,$f0
mfc1 $6,$f1
lw $2,%got($LC0)($28)
nop
addiu $4,$2,%lo($LC0)
lw $2,%call16(printf)($28)
nop
move $25,$2
.reloc 1f,R_MIPS_JALR,printf
1: jalr $25
nop
lw $28,16($fp)
move $2,$0
move $sp,$fp
lw $31,76($sp)
lw $fp,72($sp)
addiu $sp,$sp,80
j $31
nop
.set macro
.set reorder
.end main
.size main, .-main
.ident "GCC: (Ubuntu 5.4.0-6ubuntu1~16.04.9) 5.4.0 20160609"
This is the part I could not understand:
lw $2,68($fp)
nop
mtc1 $2,$f0
nop
cvt.d.w $f0,$f0
mov.d $f12,$f0
lw $2,%call16(sqrt)($28)
nop
move $25,$2
.reloc 1f,R_MIPS_JALR,sqrt
1: jalr $25
nop
lw $28,16($fp)
cvt.s.d $f0,$f0
swc1 $f0,28($fp)
lwc1 $f0,28($fp)
nop
cvt.d.s $f0,$f0
mfc1 $7,$f0
mfc1 $6,$f1
lw $2,%got($LC0)($28)
nop
addiu $4,$2,%lo($LC0)
lw $2,%call16(printf)($28)
nop
move $25,$2
.reloc 1f,R_MIPS_JALR,printf
1: jalr $25
nop
lw $28,16($fp)
move $2,$0
move $sp,$fp
lw $31,76($sp)
lw $fp,72($sp)
addiu $sp,$sp,80
j $31
nop
.set macro
.set reorder
.end main
.size main, .-main
.ident "GCC: (Ubuntu 5.4.0-6ubuntu1~16.04.9) 5.4.0 20160609"
What does .reloc mean?
How does %call16 work?
What is the corresponding value of the variable fibSquared?
How the functions "printf" and "sqrt" are called and executed?

Answer 1

I think you take the same class I do because I have the same homework, so I am going to share what I understood myself, it may help.
For %call16, if you noticed between the brackets, the name that the C code uses from the included files such as printf from stdio.h, sqrt from math file, so it's clearly calling them so he can use them.
About the .reloc, I have noticed that they come after %call16 and the called function executed and is for relocating the memory, so I think it reorders the included files from any damage that may have happened.
The fibSquared value is located in mem[-12] (SW $3,8($2))$2=-20, then it's moved to $f0 by(mtc1 $2,$f0). In the instruction, the second par is the destination($f0). It's converted from int to float in(cvt.d.w $f0,$f0).
The sqrt executed in this part of the code:
lw $2,%call16(sqrt)($28) \\ in this particular
nop
move $25,$2
.reloc 1f,R_MIPS_JALR,sqrt
1: jalr $25
$25 at the beginning was chosen for cpload (.cpload $25)
** printf :
lw $2,%got($LC0)($28)
nop
addiu $4,$2,%lo($LC0)
lw $2,%call16(printf)($28)\\ in this particular
nop
move $25,$2
.reloc 1f,R_MIPS_JALR,printf
1: jalr $25