如何将无符号长整型转换为所需格式的字符串 [英] How Do I Convert Unsigned Long Long To A String In The Desired Format

问题描述

我正在尝试转换无符号长long val = 0x1234567890123456;到一个字符串0x1234567890123456，但我不能够转换它，即使我使用％llu作为格式说明符与代码部分提供的代码。

但我能够转换unsigned int val = 0x1234567A;到0x1234567A。

你能否告诉我要做到这一点我必须做些什么？感谢您的帮助。

我想转换一个无符号长long val = 0x1234567890123456;使用以下代码到字符串0x1234567890123456。

I am trying to convert an unsigned long long val = 0x1234567890123456; to a string "0x1234567890123456" but i am not able to convert it even if i use "%llu" as format specifier with the code provided in code-section .
But I am able to convert an unsigned int val = 0x1234567A; to "0x1234567A".
Could you please let me know what i have to do to achieve this ?? Appreciate your help.
I want to convert an unsigned long long val = 0x1234567890123456; to a string "0x1234567890123456" with the below code.

#include <stdio.h>      /* printf, NULL */
#include <stdlib.h>     /* strtoul */
#include<conio.h>
#include<string.h>
#include<stdlib.h>
#include<wchar.h>
#include<iostream>
using namespace std;
int main ()
{
	//unsigned long long  val = 0x1234567890123456;		
	 unsigned int  val = 0x1234567A;		

	char result[255] = {0};	
	char formatStr[20] = {0};
	char typeMod[2] = {0};
	unsigned char precision=16;

	sprintf(typeMod, "X");
	char precisionMod[4] = {0};

	if (precision > 0)
		_snprintf(precisionMod, sizeof(precisionMod), ".%u", precision);

	sprintf(formatStr, "%%%s%s%s", precisionMod, "", typeMod);

	char* output = result;

	sprintf(output, "0x");
	output += 2;

	sprintf(output, formatStr, val);  
	puts(result);
	getch();
	return 0;
}

推荐答案

使用Microsoft和GNU的C标准库，64位数字有不同的类型前缀。在为多个平台构建时，您可以使用：

There are different type prefixes for 64-bit numbers with the C standard libraries from Microsoft and GNU. When building for multiple platforms, you may use:

unsigned long long  val = 0x1234567890123456;
#ifdef _MSC_VER // Using Visual C/C++
sprintf(output, "%#16I64X", val); 
#else 
sprintf(output, "%#16llX", val); 
#endif

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