gpt4 book ai didi

C++ 连接字符串的最快方法(最佳方法)

转载 作者:行者123 更新时间:2023-11-30 05:38:32 25 4
gpt4 key购买 nike

我正在制作一个必须提供非常大的报告的应用程序,我遇到了过程中最慢的部分是当它必须创建最终的 JSON 字符串时,所以我尝试了不同的方法来加快速度,看看哪个是最快的

我从对字符串的 +=+ 操作开始。然后我用 += 改变了所有的 + 并且它改进了一点。然后我尝试了一些更大胆的方法,它使用 char * buffer 并自己管理连接,甚至制作了我自己的 INTFLOAT 到字符串功能。它缩短了 30% 的时间,但这个过程仍然需要大约 15 段的大报告,我想将时间减少到最少,所以我会在这里粘贴我的代码,看看是否有任何关于如何改进性能的建议.

char * itoa2(UINT num, char * str)
{
if (num == 0)
{
*str++ = '0';
return str;
}

// Process individual digits
UCHAR nsiz = (UCHAR)log10(num) + 1;
str += nsiz;
while (num != 0)
{
*str-- = (num % 10) + '0';
num = num/10;
}
return str + nsiz + 1;
}

char * ftoa2(float num, char * str) {
int intg = (UINT)num;
int dec = (UINT)((num - intg) * 1000);



// Process individual digits
UCHAR nsiz = (UCHAR)log10(intg) + 1;
str += nsiz - 1;
if (dec > 0)
{
nsiz += (UCHAR)log10(dec) + 2;
str += nsiz - 1;
while (dec != 0) {
*str-- = (dec % 10) + '0';
dec = dec / 10;
}
*str-- = '.';
}

if (intg == 0)
{
*str-- = '0';
} else {
while (intg != 0) {
*str-- = (intg % 10) + '0';
intg = intg / 10;
}
}


return str + nsiz + 1;
}

void Formatter::jsonFormat2() {

char * result = new char[1024 * 1024 * 1024]; //TODO: Use small buffers instead of this

char * pos;

string tmp;

pos = result;

(*pos++) = '[';

const char * key;
int keyPos;
const Reporter::Entry * entry;
set<Field *, Field::cmpOrder> * grouped = params->getGrouped();

bool byDay, byMonth, byYear;
time_t ts = 0;
struct tm * lt;

char type;

FieldData::fieldVariant fkey(0, nullptr);
FieldData::fieldVariant * fvar;

byDay = params->getByDay();
byMonth = params->getByMonth();
byYear = params->getByYear();


for (auto& it : *report) {
key = it.first;
entry = it.second;
keyPos = 0;

(*pos++) = '{';

if (byDay || byMonth || byYear) {
ts = *(time_t *)key;
keyPos += sizeof(time_t);
lt = gmtime(&ts);
}

if (byDay) {
memcpy(pos, "\"day\":\"", 7);
pos+=7;
pos = itoa2(lt->tm_year + 1900, pos);
(*pos++) = '-';
if (lt->tm_mon+1 < 10) (*pos++) = '0';
pos = itoa2(lt->tm_mon+1, pos);
(*pos++) = '-';
if (lt->tm_mday < 10) (*pos++) = '0';
pos = itoa2(lt->tm_mday, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"wday\":\"", 8);
pos+=8;

memcpy(pos, &weekDayC[lt->tm_wday * 3], 3);
pos+=3;

(*pos++) = '\"'; (*pos++) = ',';
}
if (byMonth) {
memcpy(pos, "\"month\":\"", 9);
pos+=9;
pos = itoa2(lt->tm_year + 1900, pos);
(*pos++) = '-';
if (lt->tm_mon+1 < 10) (*pos++) = '0';
pos = itoa2(lt->tm_mon, pos);
(*pos++) = '\"'; (*pos++) = ',';
}
if (byYear) {
memcpy(pos, "\"year\":\"", 8);
pos+=9;
pos = itoa2(lt->tm_year + 1900, pos);
(*pos++) = '\"'; (*pos++) = ',';
}


for (auto& field : *grouped) {
(*pos++) = '\"';

tmp = field->getName();
memcpy(pos, tmp.c_str(), tmp.length());
pos+=tmp.length();
memcpy(pos, "\":\"", 3);
pos+=3;

type = field->getType();
if (type == INT8) {
pos = itoa2(*(key + keyPos), pos);
} else if (type == INT16) {
pos = itoa2(*(USHORT *)(key + keyPos), pos);
} else if (type == INT32) {
pos = itoa2(*(UINT *)(key + keyPos), pos);
} else if (type == INT64) {
pos = itoa2(*(ULONG *)(key + keyPos), pos);
} else if (type == INT128) {
pos = itoa2(*(UXLONG *)(key + keyPos), pos);
} else if (type == CHAR2) {
memcpy(pos, key + keyPos, 2);
pos+=2;
}

(*pos++) = '\"'; (*pos++) = ',';


for (auto& fieldData : *field->getFieldData()) {
fkey.type = field->getType();

if (fkey.type == CHAR2) {
fkey.data = new string((char *) (key + keyPos), 2);
} else
fkey.data = (void *)key + keyPos;

fvar = fieldData->find(&fkey);

(*pos++) = '\"';

tmp = fieldData->colName;
memcpy(pos, tmp.c_str(), tmp.length());
pos+=tmp.length();
memcpy(pos, "\":\"", 3);
pos+=3;
if (fvar != nullptr) tmp = toString(fvar); else tmp = "null";
memcpy(pos, tmp.c_str(), tmp.length());
pos+=tmp.length();
memcpy(pos, "\",", 2);
pos+=2;

}
keyPos += field->getBytes();
}

memcpy(pos, "\"imps\":\"", 8);
pos+=8;
pos = itoa2(entry->imps, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"impsn\":\"", 9);
pos+=9;
pos = itoa2(entry->impsn, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"impsu\":\"", 9);
pos+=9;
pos = itoa2(entry->impsu, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"clicks\":\"", 10);
pos+=10;
pos = itoa2(entry->clicks, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"clicksu\":\"", 11);
pos+=11;
pos = itoa2(entry->clicksu, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"revenue\":\"", 11);
pos+=11;
pos = ftoa2(entry->revenue, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"revenue_2\":\"", 13);
pos+=13;
pos = ftoa2(entry->revenue_2, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"revenue_3\":\"", 13);
pos+=13;
pos = ftoa2(entry->revenue_3, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"convs\":\"", 9);
pos+=9;
pos = itoa2(entry->convs, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"convs_2\":\"", 11);
pos+=11;
pos = itoa2(entry->convs_2, pos);
(*pos++) = '\"'; (*pos++) = ',';

memcpy(pos, "\"convs_3\":\"", 11);
pos+=11;
pos = itoa2(entry->convs_3, pos);
(*pos++) = '\"'; (*pos++) = ',';

(*pos++) = '}'; (*pos++) = ',';
pos+=3;
}
pos--;

(*pos++) = ']';
(*pos++) = '\0';

if (params->getVerbose()) cout << string(result, pos - result);
}

最佳答案

同样的方法,Formatter::jsonFormat2() , 你有

char * result = new char[1024 * 1024 * 1024];

在顶部,

if (params->getVerbose()) cout << string(result, pos - result);

在底部。

这意味着,您不仅每次调用此方法都会有 1GB 的内存泄漏,如果getVerbose(),您还白做了很多工作。恰好返回 false(您可以在一开始就检查一下)。

此外,如果您打算将所有内容写入std::cout (或与此相关的任何流),考虑使用流插入语法(<< 运算符)直接编写各种片段。

关于C++ 连接字符串的最快方法(最佳方法),我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32705130/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com