请教unsigned long转换为char*的问题

[WZCNet]

unsigned long long m_actorid;

m_actorid = 1020301000006;
char tmpChar[8];

memcpy(tmpChar, &m_actorid, sizeof(unsigned long long));
cout << tmpChar;

这段代码打印的结果不对。
请问如何能使tmpChar的值为1020301000006，并且能读取使用。

[jixingzhong]

unsigned long long m_actorid;

m_actorid = 1020301000006;
char tmpChar[16]=""; //8空间不足吧？

//memcpy(tmpChar, &m_actorid, sizeof(unsigned long long));
sprintf(tmpChar, "%lu", m_actorid); //sprintf格式化字符串
cout << tmpChar;

[jixingzhong]

函数名称: sprintf
函数原型: int sprintf (char *buffer,const char *format,… )
函数功能: buffer写入的缓冲区
函数返回: 写入的字符数（不包括终止符）
参数说明:
所属文件: <stdio.h>

#include <stdio.h>
#include <math.h>
int main()
{
char buffer[80];
sprintf(buffer,"An approximation of Pi is %f",M_PI);
puts(buffer);
return 0;
}

[jixingzhong]

参考函数：

函数名称: ltoa
函数原型: char *ltoa(long value,char *string,int radix)
函数功能: 将长整形数转换为等价的字符串
函数返回: 指向string的指针
参数说明: value-转换的长整形数，radix-数制(如10表示十进制）
所属文件: <stdlib.h>

#include <stdlib.h>
#include <stdio.h>
int main()
{
long number=12345;
char string[25];
itoa(number,string,10);
printf("integer=%d string=%s",number,string);
return 0;
}

[xiaoyisnail]

[Quote=引用 1 楼 jixingzhong 的回复:]
unsigned long long m_actorid;

m_actorid = 1020301000006;
char tmpChar[16]="";  //8空间不足吧？

//memcpy(tmpChar, &m_actorid, sizeof(unsigned long long));
sprintf(tmpChar, "%lu", m_actorid);  //sprintf格式化字符串
cout < < tmpChar;
[/Quote]

unsigned long long太大了，会被截断的

[xiaoyisnail]

上面的只适用于unsinged long，不适用于unsinged long long

[nineforever]

sprintf(buffer,"%llu",m_actorid);
当然你的buffer要足够大

[waizqfor]

[Quote=引用楼主 WZCNet 的帖子:]
unsigned long long m_actorid;

m_actorid = 1020301000006;
char tmpChar[8];

memcpy(tmpChar, &m_actorid, sizeof(unsigned long long));
cout < < tmpChar;

这段代码打印的结果不对。
请问如何能使tmpChar的值为1020301000006，并且能读取使用。
[/Quote]
数组好象太小了吧 不能存储那么大的数

[lanmaozi]

如果支持__int64

	__int64 int64a = 18446744073709551615;
	char szBuf[21];
	sprintf(szBuf,"%I64u", int64a);
	printf("%s\n", szBuf);

[xiaoyisnail]

测试了一下，发现格式化串用"%llu"时ok

#include <stdio.h> 
#include <iostream>

using namespace std;

int main() 
{ 
    unsigned long long m_actorid;

	m_actorid = 1020301000006;
	char tmpChar[16] = {0};

	sprintf(tmpChar, "%llu", m_actorid);  
	cout << tmpChar<<endl;

	return 0;
} 

[fallening]

[Quote=引用楼主 WZCNet 的帖子:]
unsigned long long m_actorid;

m_actorid = 1020301000006;
char tmpChar[8];

memcpy(tmpChar, &m_actorid, sizeof(unsigned long long));
cout < < tmpChar;

这段代码打印的结果不对。
请问如何能使tmpChar的值为1020301000006，并且能读取使用。
[/Quote]

string s = boost::lexical_cast<string>(m_actorid);
ans = s.c_str();

[waizqfor]

[Quote=引用 9 楼 xiaoyisnail 的回复:]
测试了一下，发现格式化串用"%llu"时ok

C/C++ code
#include <stdio.h>
#include <iostream>

using namespace std;

int main()
{
unsigned long long m_actorid;

m_actorid = 1020301000006;
char tmpChar[16] = {0};

sprintf(tmpChar, "%llu", m_actorid);
cout << tmpChar<<endl;

return 0;
}
[/Quote]
是因为unsigned long long 太大原因被截断了
%llu能正确表达

[waizqfor]

[Quote=引用 9 楼 xiaoyisnail 的回复:]
测试了一下，发现格式化串用"%llu"时ok

C/C++ code
#include <stdio.h>
#include <iostream>

using namespace std;

int main()
{
unsigned long long m_actorid;

m_actorid = 1020301000006;
char tmpChar[16] = {0};

sprintf(tmpChar, "%llu", m_actorid);
cout << tmpChar<<endl;

return 0;
}
[/Quote]
是因为unsigned long long 太大原因被截断了
%llu能正确表达

[flydream1980]

[Quote=引用 9 楼 xiaoyisnail 的回复:]
测试了一下，发现格式化串用"%llu"时ok

C/C++ code
#include <stdio.h>
#include <iostream>

using namespace std;

int main()
{
unsigned long long m_actorid;

m_actorid = 1020301000006;
char tmpChar[16] = {0};

sprintf(tmpChar, "%llu", m_actorid);
cout << tmpChar<<endl;

return 0;
}
[/Quote]

为什么结果是 2393750854， 貌似和楼主说的不一样吧，不是应该输出1020301000006 ??

[caremsi]

学习了

[fenglinqiu]

学习了