I have to round off a float to decimal. After rounding off, I should convert this number to hexadecimal. I think I got the round off part okay with round()
我必须将浮点数舍入为十进制。四舍五入后，我应该将此数字转换为十六进制。我认为圆形部分可以得到圆形部分（）
Is there a way to convert a decimal to hexadecimal in C, and store it into a part of an array? I'm thinking of the concept on how printf() converts the decimal to hex.
有没有办法在C中将十进制转换为十六进制，并将其存储到数组的一部分？我正在考虑printf（）如何将十进制转换为十六进制的概念。
What I have in mind is something like this:
我的想法是这样的：
float k = 10.123;
int a;
unsigned char var_store[1];
unsigned char array_t[3];
array_t[0] = 0x01;
array_t[1] = 0x04;
a = round(k);
var_store[0] = sprintf("%x",a);
array_t[2] = var_store[0];
but I'm having a
但我有一个
warning passing argument 2 of 'sprintf' makes pointer from integer without a cast
警告传递'sprintf'的参数2使得整数指针没有强制转换
I'm not sure if this is the way to do it. But I think this is relatively straight forward. Thanks
我不确定这是不是这样做的。但我认为这是相对直接的。谢谢
2 个解决方案
#1
2
People tend to get very confused with the term "hexadecimal". It should mean "the number as a human-readable ascii string with digits 0-F", but because raw binary data is typically presented in hex, people miuse it to mean the binary data itself. Whilst of course you can write a function that converts a decimal number, expressed as a string, to a hexadecimal number, expressed as another string, it's fiddly and, except as a learning exercise, pointless thing to do. sprintf converts C variables to human-readable strings for you. To get a decimal, pass "%d", to get hex, pass "%x". You also need to pass a destination buffer, like this.
人们倾向于对术语“十六进制”感到困惑。它应该表示“数字为人类可读的ascii字符串，数字为0-F”，但由于原始二进制数据通常以十六进制表示，人们将其称为二进制数据本身。当然，您可以编写一个函数，将表示为字符串的十进制数转换为十六进制数，表示为另一个字符串，它是繁琐的，除了作为学习练习外，无意义的事情要做。 sprintf为您将C变量转换为人类可读的字符串。要获得小数，请传递“％d”，以获取十六进制，传递“％x”。您还需要传递目标缓冲区，如下所示。
char destination[256];
int a = 123;
sprintf(destination, "number is decimal %d hex %x", a, a);
#2
0
I did not recollect any library function.
我没有回忆任何库函数。
But the traditional mathematical way is below. I you want you can create a user defined function.
但传统的数学方法如下。我希望你能创建一个用户定义的函数。
#include <iostream>
using namespace std;
int main()
{
long int decimalNumber = 2567888;
char hexadecimalNumber[100];
int temp;
int i =1;
while(decimalNumber!=0)
{
temp = decimalNumber % 16;
//To convert integer into character
if( temp < 10)
temp =temp + 48;
else
temp = temp + 55;
hexadecimalNumber[i++]= temp;
decimalNumber = decimalNumber / 16;
}
for(int j = i -1 ;j> 0;j--)
cout<<hexadecimalNumber[j];
}
#1
2
People tend to get very confused with the term "hexadecimal". It should mean "the number as a human-readable ascii string with digits 0-F", but because raw binary data is typically presented in hex, people miuse it to mean the binary data itself. Whilst of course you can write a function that converts a decimal number, expressed as a string, to a hexadecimal number, expressed as another string, it's fiddly and, except as a learning exercise, pointless thing to do. sprintf converts C variables to human-readable strings for you. To get a decimal, pass "%d", to get hex, pass "%x". You also need to pass a destination buffer, like this.
人们倾向于对术语“十六进制”感到困惑。它应该表示“数字为人类可读的ascii字符串，数字为0-F”，但由于原始二进制数据通常以十六进制表示，人们将其称为二进制数据本身。当然，您可以编写一个函数，将表示为字符串的十进制数转换为十六进制数，表示为另一个字符串，它是繁琐的，除了作为学习练习外，无意义的事情要做。 sprintf为您将C变量转换为人类可读的字符串。要获得小数，请传递“％d”，以获取十六进制，传递“％x”。您还需要传递目标缓冲区，如下所示。
char destination[256];
int a = 123;
sprintf(destination, "number is decimal %d hex %x", a, a);
#2
0
I did not recollect any library function.
我没有回忆任何库函数。
But the traditional mathematical way is below. I you want you can create a user defined function.
但传统的数学方法如下。我希望你能创建一个用户定义的函数。
#include <iostream>
using namespace std;
int main()
{
long int decimalNumber = 2567888;
char hexadecimalNumber[100];
int temp;
int i =1;
while(decimalNumber!=0)
{
temp = decimalNumber % 16;
//To convert integer into character
if( temp < 10)
temp =temp + 48;
else
temp = temp + 55;
hexadecimalNumber[i++]= temp;
decimalNumber = decimalNumber / 16;
}
for(int j = i -1 ;j> 0;j--)
cout<<hexadecimalNumber[j];
}
