C++代碼構造和c的析構函數組件++
#include <cstdio>
#include <cstdlib>
struct trivialStruct
{
trivialStruct();
~trivialStruct();
int *a;
float *b;
float *c;
};
trivialStruct::trivialStruct() : a((int*)malloc(sizeof(int))), b((float*)malloc(sizeof(float))), c((float*)malloc(sizeof(float)))
{
*a = 100;
*b = 200;
*c = 300;
}
trivialStruct::~trivialStruct()
{
free(a);
free(b);
free(c);
a = nullptr;
b = nullptr;
c = nullptr;
}
int main()
{
trivialStruct A;
printf("%d, %f, %f", *A.a, *A.b, *A.c);
return 0;
}
組件
.section __TEXT,__text,regular,pure_instructions
.globl __ZN13trivialStructC1Ev
.align 4, 0x90
__ZN13trivialStructC1Ev: ## @_ZN13trivialStructC1Ev
.cfi_startproc
## BB#0: ## %entry
push RBP
Ltmp3:
.cfi_def_cfa_offset 16
Ltmp4:
.cfi_offset rbp, -16
mov RBP, RSP
Ltmp5:
.cfi_def_cfa_register rbp
push R15
push R14
push RBX
push RAX
Ltmp6:
.cfi_offset rbx, -40
Ltmp7:
.cfi_offset r14, -32
Ltmp8:
.cfi_offset r15, -24
mov RBX, RDI
mov EDI, 4
call _malloc
mov R14, RAX
mov QWORD PTR [RBX], R14
mov EDI, 4
call _malloc
mov R15, RAX
mov QWORD PTR [RBX + 8], R15
mov EDI, 4
call _malloc
mov QWORD PTR [RBX + 16], RAX
mov DWORD PTR [R14], 100
mov DWORD PTR [R15], 1128792064
mov DWORD PTR [RAX], 1133903872
add RSP, 8
pop RBX
pop R14
pop R15
pop RBP
ret
.cfi_endproc
.globl __ZN13trivialStructC2Ev
.align 4, 0x90
__ZN13trivialStructC2Ev: ## @_ZN13trivialStructC2Ev
.cfi_startproc
## BB#0: ## %entry
push RBP
Ltmp12:
.cfi_def_cfa_offset 16
Ltmp13:
.cfi_offset rbp, -16
mov RBP, RSP
Ltmp14:
.cfi_def_cfa_register rbp
push R15
push R14
push RBX
push RAX
Ltmp15:
.cfi_offset rbx, -40
Ltmp16:
.cfi_offset r14, -32
Ltmp17:
.cfi_offset r15, -24
mov RBX, RDI
mov EDI, 4
call _malloc
mov R14, RAX
mov QWORD PTR [RBX], R14
mov EDI, 4
call _malloc
mov R15, RAX
mov QWORD PTR [RBX + 8], R15
mov EDI, 4
call _malloc
mov QWORD PTR [RBX + 16], RAX
mov DWORD PTR [R14], 100
mov DWORD PTR [R15], 1128792064
mov DWORD PTR [RAX], 1133903872
add RSP, 8
pop RBX
pop R14
pop R15
pop RBP
ret
.cfi_endproc
.globl __ZN13trivialStructD1Ev
.align 4, 0x90
__ZN13trivialStructD1Ev: ## @_ZN13trivialStructD1Ev
.cfi_startproc
## BB#0: ## %entry
push RBP
Ltmp21:
.cfi_def_cfa_offset 16
Ltmp22:
.cfi_offset rbp, -16
mov RBP, RSP
Ltmp23:
.cfi_def_cfa_register rbp
push RBX
push RAX
Ltmp24:
.cfi_offset rbx, -24
mov RBX, RDI
mov RDI, QWORD PTR [RBX]
call _free
mov RDI, QWORD PTR [RBX + 8]
call _free
mov RDI, QWORD PTR [RBX + 16]
call _free
mov QWORD PTR [RBX + 16], 0
mov QWORD PTR [RBX + 8], 0
mov QWORD PTR [RBX], 0
add RSP, 8
pop RBX
pop RBP
ret
.cfi_endproc
.globl __ZN13trivialStructD2Ev
.align 4, 0x90
__ZN13trivialStructD2Ev: ## @_ZN13trivialStructD2Ev
.cfi_startproc
## BB#0: ## %entry
push RBP
Ltmp28:
.cfi_def_cfa_offset 16
Ltmp29:
.cfi_offset rbp, -16
mov RBP, RSP
Ltmp30:
.cfi_def_cfa_register rbp
push RBX
push RAX
Ltmp31:
.cfi_offset rbx, -24
mov RBX, RDI
mov RDI, QWORD PTR [RBX]
call _free
mov RDI, QWORD PTR [RBX + 8]
call _free
mov RDI, QWORD PTR [RBX + 16]
call _free
mov QWORD PTR [RBX + 16], 0
mov QWORD PTR [RBX + 8], 0
mov QWORD PTR [RBX], 0
add RSP, 8
pop RBX
pop RBP
ret
.cfi_endproc
.section __TEXT,__literal8,8byte_literals
.align 3
LCPI4_0:
.quad 4641240890982006784 ## double 200
LCPI4_1:
.quad 4643985272004935680 ## double 300
.section __TEXT,__text,regular,pure_instructions
.globl _main
.align 4, 0x90
_main: ## @main
.cfi_startproc
## BB#0: ## %entry
push RBP
Ltmp34:
.cfi_def_cfa_offset 16
Ltmp35:
.cfi_offset rbp, -16
mov RBP, RSP
Ltmp36:
.cfi_def_cfa_register rbp
lea RDI, QWORD PTR [RIP + L_.str]
movsd XMM0, QWORD PTR [RIP + LCPI4_0]
movsd XMM1, QWORD PTR [RIP + LCPI4_1]
mov ESI, 100
mov AL, 2
call _printf
xor EAX, EAX
pop RBP
ret
.cfi_endproc
.section __TEXT,__cstring,cstring_literals
L_.str: ## @.str
.asciz "%d, %f, %f"
.subsections_via_symbols
命令 鐺++ -S -O2 -std = C++ 11 -mllvm --x86-ASM-語法=英特爾-fno-例外的main.cpp
正如你所看到的,有碼的兩個部分是相同的(構造函數和析構函數)
- __ZN13trivialStructC1Ev:## @ _ZN13trivialStructC1Ev
- __ZN13trivialStructC2Ev:## @ _ZN13trivialStructC2Ev
- __ZN13trivialStructD1Ev:## @ _ZN13trivialStructD1Ev
- __ZN13trivialStructD2Ev:## @ _ZN13trivialStructD2Ev
我不知道爲什麼編譯器生成兩部分代碼,但不只是一個? 我並不熟悉程序集,但看起來像這樣只是使代碼變得更胖(也許更慢)。
嘗試優化代碼大小(將'-Os'傳遞給g ++)。此外,較少的代碼並不總是更好 - 分支可能非常昂貴。 –
請注意,代碼不會*慢*。如果不使用構造函數,鏈接器可能會刪除符號(請閱讀鏈接器的文檔),即使它們未被刪除,它們也會佔用一些內存空間,但不會被評估(即不會導致其他代碼被從緩存中刪除)。 –