打印性能优化[缓冲区大小角度]

Posted 2022-02-03 BHY_

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问题背景：

https://blog.csdn.net/qq_15947787/article/details/119001748?spm=1001.2014.3001.5501

问题分析角度：

针对YY个字段信息，每个字段要打印3列信息，按较大值估计一行50个字节，最多可达 ?K左右。从理论上将，超出缓冲区将会引起多次的write操作。

对于超过缓冲区大小输出分析：

打印代码

Tick差

分析

string str = "";

for (int i = 0; i < 5000; i++)

for (int j = 0; j < 2000; j++)

str += "1";

fout << str << "\\n";

str.clear();

小于2K

144714

145785

146577

17:10:43.486858 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 2000], 2) = 2001 <0.000028>

17:10:43.486967 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 2000], 2) = 2001 <0.000017>

17:10:43.487065 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 2000], 2) = 2001 <0.000021>

17:10:43.487170 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 2000], 2) = 2001 <0.000016>

17:10:43.487266 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 2000], 2) = 2001 <0.000021>

这里str+=”1”测试为”12”和”1\\n”时性能没有明显差异

string str = "";

for (int i = 0; i < 2000; i++)

for (int j = 0; j < 5000; j++)

str += "1";

fout << str << "\\n";

str.clear();

超4K

139111

136928

138226

17:13:12.839220 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 5000], 2) = 5001 <0.000018>

17:13:12.839329 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 5000], 2) = 5001 <0.000018>

17:13:12.839436 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 5000], 2) = 5001 <0.000018>

17:13:12.839545 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 5000], 2) = 5001 <0.000016>

17:13:12.839652 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 5000], 2) = 5001 <0.000021>

string str = "";

for (int i = 0; i < 1000; i++)

for (int j = 0; j < 10000; j++)

str += "1";

fout << str << "\\n";

str.clear();

超8K

131784

137378

131725

17:15:14.229337 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 10000], 2) = 10001 <0.000028>

17:15:14.229515 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 10000], 2) = 10001 <0.000035>

17:15:14.229700 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 10000], 2) = 10001 <0.000021>

17:15:14.229871 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 10000], 2) = 10001 <0.000021>

17:15:14.230042 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 10000], 2) = 10001 <0.000025>

使用string拼接，统一输出，调用writev，string直接传地址，大小为10000

string str = "";

for (int i = 0; i < 1000; i++)

for (int j = 0; j < 10000; j++)

str += "1";

fout << str << endl;

str.clear();

133347

135463

135448

17:17:26.164393 writev(3, ["", 0, "11111111111111111111111111111111"..., 10000], 2) = 10000 <0.000023>

17:17:26.164446 write(3, "\\n", 1) = 1 <0.000012>

17:17:26.164606 writev(3, ["", 0, "11111111111111111111111111111111"..., 10000], 2) = 10000 <0.000026>

17:17:26.164662 write(3, "\\n", 1) = 1 <0.000013>

17:17:26.164822 writev(3, ["", 0, "11111111111111111111111111111111"..., 10000], 2) = 10000 <0.000022>

17:17:26.164874 write(3, "\\n", 1) = 1 <0.000012>

string输出时的换行，依然由write完成

for (int i = 0; i < 1000; i++)

for (int j = 0; j < 10000; j++)

fout << "1";

fout << "\\n";

465279

467595

471035

16:00:40.941887 writev(3, ["11111111111111111111111111111111"..., 8190, "1", 1], 2) = 8191 <0.000021>

16:00:40.942307 writev(3, ["11111111111111111111111111111111"..., 8190, "1", 1], 2) = 8191 <0.000020>

16:00:40.942727 writev(3, ["11111111111111111111\\n11111111111"..., 8190, "1", 1], 2) = 8191 <0.000022>

16:00:40.943148 writev(3, ["11111111111111111111111111111111"..., 8190, "1", 1], 2) = 8191 <0.000023>

16:00:40.943570 writev(3, ["11111111111111111111111111111111"..., 8190, "1", 1], 2) = 8191 <0.000022>

直接写os，默认缓冲区8k，包括换行在内持续输出

for (int i = 0; i < 1000; i++)

for (int j = 0; j < 10000; j++)

fout << "1";

fout << endl;

467999

467723

468518

17:20:54.810326 writev(3, ["11111111111111111111111111111111"..., 8190, "1", 1], 2) = 8191 <0.000025>

17:20:54.810462 write(3, "11111111111111111111111111111111"..., 1810) = 1810 <0.000013>

17:20:54.810872 writev(3, ["11111111111111111111111111111111"..., 8190, "1", 1], 2) = 8191 <0.000022>

17:20:54.811007 write(3, "11111111111111111111111111111111"..., 1810) = 1810 <0.000018>

17:20:54.811423 writev(3, ["11111111111111111111111111111111"..., 8190, "1", 1], 2) = 8191 <0.000023>

17:20:54.811558 write(3, "11111111111111111111111111111111"..., 1810) = 1810 <0.000012>

直接写os，10000次后写入endl，则前8192次直接写，后1810次write写

for (int i = 0; i < 100000; i++)

fout << "1" <<endl;

156273

157962

155850

17:45:54.054019 write(3, "1\\n", 2) = 2 <0.000012>

17:45:54.054058 write(3, "1\\n", 2) = 2 <0.000012>

17:45:54.054098 write(3, "1\\n", 2) = 2 <0.000012>

17:45:54.054137 write(3, "1\\n", 2) = 2 <0.000011>

17:45:54.054177 write(3, "1\\n", 2) = 2 <0.000012>

for (int i = 0; i < 100000; i++)

fout << "1" << "\\n";

14744

14365

14147

17:46:55.646126 writev(3, ["\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1"..., 8190, "\\n", 1], 2) = 8191 <0.000022>

17:46:55.646544 writev(3, ["1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 8190, "1", 1], 2) = 8191 <0.000021>

17:46:55.646960 writev(3, ["\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1"..., 8190, "\\n", 1], 2) = 8191 <0.000021>

17:46:55.647375 writev(3, ["1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 8190, "1", 1], 2) = 8191 <0.000024>

17:46:55.647793 writev(3, ["\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1"..., 8190, "\\n", 1], 2) = 8191 <0.000021>

for (int i = 0; i < 100000; i++)

fout << "1\\n";

7986

5187

5084

17:43:24.933016 writev(3, ["1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 8190, "1\\n", 2], 2) = 8192 <0.000019>

17:43:24.933250 writev(3, ["1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 8190, "1\\n", 2], 2) = 8192 <0.000019>

17:43:24.933485 writev(3, ["1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 8190, "1\\n", 2], 2) = 8192 <0.000020>

17:43:24.933720 writev(3, ["1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 8190, "1\\n", 2], 2) = 8192 <0.000020>

17:43:24.933954 writev(3, ["1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 8190, "1\\n", 2], 2) = 8192 <0.000019>

"1\\n"连续的优势

string str = "";

for (int i = 0; i < 100000; i++)

str += "12";

fout << str <<endl;

2770

2582

2535

15:00:55.015926 writev(3, ["", 0, "12121212121212121212121212121212"..., 200000], 2) = 200000 <0.000322>

15:00:55.016320 write(3, "\\n", 1) = 1 <0.000016>

15:00:55.018379 writev(3, ["", 0, "12121212121212121212121212121212"..., 200000], 2) = 200000 <0.000215>

15:00:55.018654 write(3, "\\n", 1) = 1 <0.000015>

15:00:55.020676 writev(3, ["", 0, "12121212121212121212121212121212"..., 200000], 2) = 200000 <0.000286>

15:00:55.021025 write(3, "\\n", 1) = 1 <0.000015>

如果拆成2次str +=x

性能略微降低

string str = "";

for (int i = 0; i < 100000; i++)

str += "1\\n ";

fout << str <<endl;

2477

2550

2445

14:59:43.635929 writev(3, ["", 0, "1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 200000], 2) = 200000 <0.000241>

14:59:43.636237 write(3, "\\n", 1) = 1 <0.000016>

14:59:43.638280 writev(3, ["", 0, "1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 200000], 2) = 200000 <0.000275>

14:59:43.638627 write(3, "\\n", 1) = 1 <0.000016>

14:59:43.640692 writev(3, ["", 0, "1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n1\\n"..., 200000], 2) = 200000 <0.000219>

14:59:43.640972 write(3, "\\n", 1) = 1 <0.000015>

\\n和普通字符同等对待

string str = "";

for (int i = 0; i < 500; i++)

for (int j = 0; j < 20000; j++)

str += "1";

fout << str << "\\n";

str.clear();

136666

137532

132687

17:24:03.514167 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 20000], 2) = 20001 <0.000034>

17:24:03.514473 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 20000], 2) = 20001 <0.000038>

17:24:03.514784 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 20000], 2) = 20001 <0.000031>

17:24:03.515089 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 20000], 2) = 20001 <0.000034>

17:24:03.515395 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 20000], 2) = 20001 <0.000032>

string str = "";

for (int i = 0; i < 50; i++)

for (int j = 0; j < 200000; j++)

str += "1";

fout << str << "\\n";

str.clear();

134238

138384

134882

17:26:17.040128 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 200000], 2) = 200001 <0.000287>

17:26:17.043042 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 200000], 2) = 200001 <0.000210>

17:26:17.045880 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 200000], 2) = 200001 <0.000197>

17:26:17.048703 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 200000], 2) = 200001 <0.000202>

17:26:17.051510 writev(3, ["\\n", 1, "11111111111111111111111111111111"..., 200000], 2) = 200001 <0.000195>

用string做字符串拼接，最后一次写入的方式中，总写入量一定，耗费的时间差异不大。

每次都fout输出方式，达到刷缓冲区条件的方式中，性能并不理想。

os多次打印输出时，"\\n"结合到前面字符串中，可提升一倍效率。

反汇编对比：

打印代码

int main()

std::ofstream fout("1.txt");

for (int j = 0; j < 10000; j++)

fout << "1" << endl;

fout.close();

return 0;

0000000000400a78 <main>:

400a78: 55 push %rbp

400a79: 48 89 e5 mov %rsp,%rbp

400a7c: 53 push %rbx

400a7d: 48 81 ec 18 02 00 00 sub $0x218,%rsp

400a84: be 20 00 00 00 mov $0x20,%esi

400a89: bf 10 00 00 00 mov $0x10,%edi

400a8e: e8 ee 00 00 00 callq 400b81 <_ZStorSt13_ios_OpenmodeS_>

400a93: 89 c2 mov %eax,%edx

400a95: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400a9c: be 55 0c 40 00 mov $0x400c55,%esi

400aa1: 48 89 c7 mov %rax,%rdi

400aa4: e8 97 fe ff ff callq 400940 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEEC1EPKcSt13_Ios_Openmode@plt>

400aa9: c7 45 ec 00 00 00 00 movl $0x0,-0x14(%rbp)

400ab0: 81 7d ec 0f 27 00 00 cmpl $0x270f,-0x14(%rbp)

400ab7: 7f 27 jg 400ae0 <main+0x68>

400ab9: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400ac0: be 5b 0c 40 00 mov $0x400c5b,%esi

400ac5: 48 89 c7 mov %rax,%rdi

400ac8: e8 33 fe ff ff callq 400900 <_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc@plt>

400acd: be 50 09 40 00 mov $0x400950,%esi

400ad2: 48 89 c7 mov %rax,%rdi

400ad5: e8 56 fe ff ff callq 400930 <_ZNSolsEPFRSoS_E@plt> // endl输出

400ada: 83 45 ec 01 addl $0x1,-0x14(%rbp)

400ade: eb d0 jmp 400ab0 <main+0x38> //0xab0位置

400ae0: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400ae7: 48 89 c7 mov %rax,%rdi

400aea: e8 31 fe ff ff callq 400920 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEE5closeEv@plt>

400aef: bb 00 00 00 00 mov $0x0,%ebx

400af4: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400afb: 48 89 c7 mov %rax,%rdi

400afe: e8 0d fe ff ff callq 400910 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEED1Ev@plt>

400b03: 89 d8 mov %ebx,%eax

400b05: eb 1d jmp 400b24 <main+0xac>

400b07: 48 89 c3 mov %rax,%rbx

400b0a: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400b11: 48 89 c7 mov %rax,%rdi

400b14: e8 f7 fd ff ff callq 400910 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEED1Ev@plt>

400b19: 48 89 d8 mov %rbx,%rax

400b1c: 48 89 c7 mov %rax,%rdi

400b1f: e8 4c fe ff ff callq 400970 <_Unwind_Resume@plt>

400b24: 48 81 c4 18 02 00 00 add $0x218,%rsp

400b2b: 5b pop %rbx

400b2c: 5d pop %rbp

400b2d: c3 retq

int main()

std::ofstream fout("1.txt");

for (int j = 0; j < 10000; j++)

fout << "1" << "\\n";

fout.close();

return 0;

0000000000400998 <main>:

400998: 55 push %rbp

400999: 48 89 e5 mov %rsp,%rbp

40099c: 53 push %rbx

40099d: 48 81 ec 18 02 00 00 sub $0x218,%rsp

4009a4: be 20 00 00 00 mov $0x20,%esi

4009a9: bf 10 00 00 00 mov $0x10,%edi

4009ae: e8 ee 00 00 00 callq 400aa1 <_ZStorSt13_Ios_OpenmodeS_>

4009b3: 89 c2 mov %eax,%edx

4009b5: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

4009bc: be 75 0b 40 00 mov $0x400b75,%esi

4009c1: 48 89 c7 mov %rax,%rdi

4009c4: e8 a7 fe ff ff callq 400870 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEEC1EPKcSt13_Ios_Openmode@plt>

4009c9: c7 45 ec 00 00 00 00 movl $0x0,-0x14(%rbp)

4009d0: 81 7d ec 0f 27 00 00 cmpl $0x270f,-0x14(%rbp)

4009d7: 7f 27 jg 400a00 <main+0x68>

4009d9: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

4009e0: be 7b 0b 40 00 mov $0x400b7b,%esi

4009e5: 48 89 c7 mov %rax,%rdi

4009e8: e8 53 fe ff ff callq 400840 <_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc@plt> 前面单独的<<”1”

4009ed: be 7d 0b 40 00 mov $0x400b7d,%esi

4009f2: 48 89 c7 mov %rax,%rdi

4009f5: e8 46 fe ff ff callq 400840 <_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc@plt> 后面单独的<<”\\n”

4009fa: 83 45 ec 01 addl $0x1,-0x14(%rbp)

4009fe: eb d0 jmp 4009d0 <main+0x38>

400a00: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400a07: 48 89 c7 mov %rax,%rdi

400a0a: e8 51 fe ff ff callq 400860 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEE5closeEv@plt>

400a0f: bb 00 00 00 00 mov $0x0,%ebx

400a14: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400a1b: 48 89 c7 mov %rax,%rdi

400a1e: e8 2d fe ff ff callq 400850 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEED1Ev@plt>

400a23: 89 d8 mov %ebx,%eax

400a25: eb 1d jmp 400a44 <main+0xac>

400a27: 48 89 c3 mov %rax,%rbx

400a2a: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400a31: 48 89 c7 mov %rax,%rdi

400a34: e8 17 fe ff ff callq 400850 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEED1Ev@plt>

400a39: 48 89 d8 mov %rbx,%rax

400a3c: 48 89 c7 mov %rax,%rdi

400a3f: e8 4c fe ff ff callq 400890 <_Unwind_Resume@plt>

400a44: 48 81 c4 18 02 00 00 add $0x218,%rsp

400a4b: 5b pop %rbx

400a4c: 5d pop %rbp

400a4d: c3 retq

int main()

std::ofstream fout("1.txt");

for (int j = 0; j < 10000; j++)

fout << "1\\n";

fout.close();

return 0;

0000000000400998 <main>:

400998: 55 push %rbp

400999: 48 89 e5 mov %rsp,%rbp

40099c: 53 push %rbx

40099d: 48 81 ec 18 02 00 00 sub $0x218,%rsp

4009a4: be 20 00 00 00 mov $0x20,%esi

4009a9: bf 10 00 00 00 mov $0x10,%edi

4009ae: e8 e1 00 00 00 callq 400a94 <_ZStorSt13_Ios_OpenmodeS_>

4009b3: 89 c2 mov %eax,%edx

4009b5: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

4009bc: be 65 0b 40 00 mov $0x400b65,%esi

4009c1: 48 89 c7 mov %rax,%rdi

4009c4: e8 a7 fe ff ff callq 400870 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEEC1EPKcSt13_Ios_Openmode@plt>

4009c9: c7 45 ec 00 00 00 00 movl $0x0,-0x14(%rbp)

4009d0: 81 7d ec 0f 27 00 00 cmpl $0x270f,-0x14(%rbp)

4009d7: 7f 1a jg 4009f3 <main+0x5b>

4009d9: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

4009e0: be 6b 0b 40 00 mov $0x400b6b,%esi

4009e5: 48 89 c7 mov %rax,%rdi

4009e8: e8 53 fe ff ff callq 400840 <_ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc@plt> <<”1\\n”输出

4009ed: 83 45 ec 01 addl $0x1,-0x14(%rbp)

4009f1: eb dd jmp 4009d0 <main+0x38>

4009f3: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

4009fa: 48 89 c7 mov %rax,%rdi

4009fd: e8 5e fe ff ff callq 400860 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEE5closeEv@plt>

400a02: bb 00 00 00 00 mov $0x0,%ebx

400a07: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400a0e: 48 89 c7 mov %rax,%rdi

400a11: e8 3a fe ff ff callq 400850 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEED1Ev@plt>

400a16: 89 d8 mov %ebx,%eax

400a18: eb 1d jmp 400a37 <main+0x9f>

400a1a: 48 89 c3 mov %rax,%rbx

400a1d: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400a24: 48 89 c7 mov %rax,%rdi

400a27: e8 24 fe ff ff callq 400850 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEED1Ev@plt>

400a2c: 48 89 d8 mov %rbx,%rax

400a2f: 48 89 c7 mov %rax,%rdi

400a32: e8 59 fe ff ff callq 400890 <_Unwind_Resume@plt>

400a37: 48 81 c4 18 02 00 00 add $0x218,%rsp

400a3e: 5b pop %rbx

400a3f: 5d pop %rbp

400a40: c3 retq

string s = "";

int main()

std::ofstream fout("1.txt");

for (int j = 0; j < 10000; j++)

s += "1\\n";

fout << s << endl;

fout.close();

return 0;

0000000000400d38 <main>:

400d38: 55 push %rbp

400d39: 48 89 e5 mov %rsp,%rbp

400d3c: 53 push %rbx

400d3d: 48 81 ec 18 02 00 00 sub $0x218,%rsp

400d44: be 20 00 00 00 mov $0x20,%esi

400d49: bf 10 00 00 00 mov $0x10,%edi

400d4e: e8 64 01 00 00 callq 400eb7 <_ZStorSt13_Ios_OpenmodeS_>

400d53: 89 c2 mov %eax,%edx

400d55: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400d5c: be 85 0f 40 00 mov $0x400f85,%esi

400d61: 48 89 c7 mov %rax,%rdi

400d64: e8 67 fe ff ff callq 400bd0 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEEC1EPKcSt13_Ios_Openmode@plt>

400d69: c7 45 ec 00 00 00 00 movl $0x0,-0x14(%rbp)

400d70: 81 7d ec 0f 27 00 00 cmpl $0x270f,-0x14(%rbp) // 循环10000次

400d77: 7f 15 jg 400d8e <main+0x56>

400d79: be 8b 0f 40 00 mov $0x400f8b,%esi

400d7e: bf e0 20 60 00 mov $0x6020e0,%edi

400d83: e8 d8 fd ff ff callq 400b60 <_ZNSt7__cxx1112basic_stringIcSt11char_traitsIcESaIcEEpLEPKc@plt>

// 调用的是std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> >::operator+=(char const*)

400d88: 83 45 ec 01 addl $0x1,-0x14(%rbp)

400d8c: eb e2 jmp 400d70 <main+0x38> // 目标0xd70，循环体处

400d8e: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400d95: be e0 20 60 00 mov $0x6020e0,%esi

400d9a: 48 89 c7 mov %rax,%rdi

400d9d: e8 7e fe ff ff callq 400c20 <_ZStlsIcSt11char_traitsIcESaIcEERSt13basic_ostreamIT_T0_ES7_RKNSt7__cxx1112basic_stringIS4_S5_T1_EE@plt>

400da2: be e0 0b 40 00 mov $0x400be0,%esi

400da7: 48 89 c7 mov %rax,%rdi

400daa: e8 11 fe ff ff callq 400bc0 <_ZNSolsEPFRSoS_E@plt>

400daf: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400db6: 48 89 c7 mov %rax,%rdi

400db9: e8 e2 fd ff ff callq 400ba0 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEE5closeEv@plt>

400dbe: bb 00 00 00 00 mov $0x0,%ebx

400dc3: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400dca: 48 89 c7 mov %rax,%rdi

400dcd: e8 be fd ff ff callq 400b90 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEED1Ev@plt>

400dd2: 89 d8 mov %ebx,%eax

400dd4: eb 1d jmp 400df3 <main+0xbb>

400dd6: 48 89 c3 mov %rax,%rbx

400dd9: 48 8d 85 e0 fd ff ff lea -0x220(%rbp),%rax

400de0: 48 89 c7 mov %rax,%rdi

400de3: e8 a8 fd ff ff callq 400b90 <_ZNSt14basic_ofstreamIcSt11char_traitsIcEED1Ev@plt>

400de8: 48 89 d8 mov %rbx,%rax

400deb: 48 89 c7 mov %rax,%rdi

400dee: e8 3d fe ff ff callq 400c30 <_Unwind_Resume@plt>

400df3: 48 81 c4 18 02 00 00 add $0x218,%rsp

400dfa: 5b pop %rbx

400dfb: 5d pop %rbp

400dfc: c3 retq

// fout << "1" << endl;

b00495022@linux-x001:~/test> c++filt _ZNSolsEPFRSoS_E

std::basic_ostream<char, std::char_traits<char> >::operator<<(std::basic_ostream<char, std::char_traits<char> >& (*)(std::basic_ostream<char, std::char_traits<char> >&))

// fout << "1" << "\\n";

b00495022@linux-x001:~/test> c++filt _ZStlsISt11char_traitsIcEERSt13basic_ostreamIcT_ES5_PKc

std::basic_ostream<char, std::char_traits<char> >& std::operator<< <std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*)

可以看出"1\\n"少次一次call，endl和“\\n”call的函数不同。

优化代码：

void functest(std::ostream &os)

std::string str = “”;

str += "xxxxx\\n";

while(flag)

str += "[";

str += "xxxyyy\\n";

str += "zzz";

str += "]\\n";

os << str << endl; // 最后确保函数走完输出到文件或者由调用者flush或者触发自动刷缓冲区条件

对于函数中打印少量内容，但换行较多，不想多增加string变量，则推荐 <<”\\n”方式。

其实将string累计字符串时，就不再涉及endl了，每次<<都会做一次call，而string方式则只做一次call，同时也避免了频繁刷缓冲区，性能上得以更大提升。

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