I have a string 2290348
I need it to display as ACF22200 (little endian)
Because the input 2290348 is passed in via a form in textbox, I have tried to read it as string (eg. this->textBox1->Text) and convert it to int (eg. Convert::ToInt32(this->textBox1->Text)).
afterwhich, i converted it to hex via ToString("x") which i manage to get 22F2AC
I appended 00 to 22F2AC and gotten 0022F2AC still as string
now i'm stuck in converting 0022F2AC to ACF22200
While still an int, you could use e.g. htonl to convert it to "network" byte order.
#include <winsock2.h>
int main()
{
unsigned int x = 0x22F2AC;
printf("x = 0x%08x\n", x);
printf("htonl(x) = 0x%08x\n", htonl(x));
return 0;
}
The program above prints:
x = 0x0022f2ac
htonl(x) = 0xacf22200
Related
I need to convert a char[] to __be16 type in linux kernel space.
I can able to convert char[] to int using kstrtoint. The same function works for __be16 too, but with warnings. Similarly any predefined functions available for __be16 too?
Example :
char x[120] = "02" to __be16 xx
__be16 actually is a network short int, so you can use htons function:
char* str = "123"; // input
int val;
__be16 nval; // output
kstrtoint(str, 10, &val);
nval = htons(val);
Also, cpu_to_be16 can be used instead of htons.
This question already has answers here:
How can I convert an int to a string in C?
(10 answers)
Closed 8 years ago.
I want to convert an int to a string without printing anything on my screen. For now I used sprintf, but this also printed the int to my screen.
Also itoa is not supported by my compiler so I can't use that either.
I assume You use ANSI C.
You cannot use itoa because it's not a standard function.
sprintf or snprintf is dedicated to that.
Since You do not want to use sprintf, make your own itoa instead:
#include <stdio.h>
char* itoa(int i, char b[]){
char const digit[] = "0123456789";
char* p = b;
if(i<0){
*p++ = '-';
i *= -1;
}
int shifter = i;
do{ //Move to where representation ends
++p;
shifter = shifter/10;
}while(shifter);
*p = '\0';
do{ //Move back, inserting digits as u go
*--p = digit[i%10];
i = i/10;
}while(i);
return b;
}
original answer: here
Some body help me regarding to the following problem
strFixFactorSide = _T("0.5");
dFixFactorSide = atof((const char *)(LPCTSTR)strFixFactorSide);
"dFixFactorSide" takes value as 0.0000;
How I will get correct value?
Use _tstof() instead of atof(), and cast CString to LPCTSTR, and leave it as such, instead of trying to get it to const char *. Forget about const char * (LPCSTR) while you're working with unicode and use only const _TCHAR * (LPCTSTR).
int _tmain(int argc, TCHAR* argv[], TCHAR* envp[])
{
int nRetCode = 0;
CString s1 = _T("123.4");
CString s2 = _T("567.8");
double v1 = _tstof((LPCTSTR)s1);
double v2 = _tstof((LPCTSTR)s2);
_tprintf(_T("%.3f"), v1 + v2);
return nRetCode;
}
and running this correctly gives the expected answer.
I think your CString strFixFactorSide is a Unicode (UTF-16) string.
If it is, the cast (const char *) only changes the pointer type, but the string it points to still remains Unicode.
atof() doesn't work with Unicode strings. If you shove L"0.5" into it, it will fetch bytes 0x30 ('0') and 0x00 (also part of UTF-16 '0'), treat that as a NUL-terminated ASCII string "0" and convert it to 0.0.
If CString strFixFactorSide is a Unicode string, you need to either first convert it to an ASCII string and then apply atof() or use a function capable of converting Unicode strings to numbers. _wtof() can be used for Unicode strings.
I am getting an int value from one of the analog pins on my Arduino. How do I concatenate this to a String and then convert the String to a char[]?
It was suggested that I try char msg[] = myString.getChars();, but I am receiving a message that getChars does not exist.
To convert and append an integer, use operator += (or member function concat):
String stringOne = "A long integer: ";
stringOne += 123456789;
To get the string as type char[], use toCharArray():
char charBuf[50];
stringOne.toCharArray(charBuf, 50)
In the example, there is only space for 49 characters (presuming it is terminated by null). You may want to make the size dynamic.
Overhead
The cost of bringing in String (it is not included if not used anywhere in the sketch), is approximately 1212 bytes of program memory (flash) and 48 bytes RAM.
This was measured using Arduino IDE version 1.8.10 (2019-09-13) for an Arduino Leonardo sketch.
Risk
There must be sufficient free RAM available. Otherwise, the result may be lockup/freeze of the application or other strange behaviour (UB).
Just as a reference, below is an example of how to convert between String and char[] with a dynamic length -
// Define
String str = "This is my string";
// Length (with one extra character for the null terminator)
int str_len = str.length() + 1;
// Prepare the character array (the buffer)
char char_array[str_len];
// Copy it over
str.toCharArray(char_array, str_len);
Yes, this is painfully obtuse for something as simple as a type conversion, but somehow it's the easiest way.
You can convert it to char* if you don't need a modifiable string by using:
(char*) yourString.c_str();
This would be very useful when you want to publish a String variable via MQTT in arduino.
None of that stuff worked. Here's a much simpler way .. the label str is the pointer to what IS an array...
String str = String(yourNumber, DEC); // Obviously .. get your int or byte into the string
str = str + '\r' + '\n'; // Add the required carriage return, optional line feed
byte str_len = str.length();
// Get the length of the whole lot .. C will kindly
// place a null at the end of the string which makes
// it by default an array[].
// The [0] element is the highest digit... so we
// have a separate place counter for the array...
byte arrayPointer = 0;
while (str_len)
{
// I was outputting the digits to the TX buffer
if ((UCSR0A & (1<<UDRE0))) // Is the TX buffer empty?
{
UDR0 = str[arrayPointer];
--str_len;
++arrayPointer;
}
}
With all the answers here, I'm surprised no one has brought up using itoa already built in.
It inserts the string representation of the integer into the given pointer.
int a = 4625;
char cStr[5]; // number of digits + 1 for null terminator
itoa(a, cStr, 10); // int value, pointer to string, base number
Or if you're unsure of the length of the string:
int b = 80085;
int len = String(b).length();
char cStr[len + 1]; // String.length() does not include the null terminator
itoa(b, cStr, 10); // or you could use String(b).toCharArray(cStr, len);
I've seen lots of answers to this, but I cannot seem to get any to work. I think I'm getting confused between variable types. I have an input from NetworkStream that is put a hex code into a String^. I need to take part of this string, convert it to a number (presumably int) so I can add some arithemetic, then output the reult on the form. The code I have so far:
String^ msg; // gets filled later, e.g. with "A55A6B0550000000FFFBDE0030C8"
String^ test;
//I have selected the relevant part of the string, e.g. 5A
test = msg->Substring(2, 2);
//I have tried many different routes to extract the numverical value of the
//substring. Below are some of them:
std::stringstream ss;
hexInt = 0;
//Works if test is string, not String^ but then I can't output it later.
ss << sscanf(test.c_str(), "%x", &hexInt);
//--------
sprintf(&hexInt, "%d", test);
//--------
//And a few others that I've deleted after they don't work at all.
//Output:
this->textBox1->AppendText("Display numerical value after a bit of math");
Any help with this would be greatly appreciated.
Chris
Does this help?
String^ hex = L"5A";
int converted = System::Convert::ToInt32(hex, 16);
The documentation for the Convert static method used is on the MSDN.
You need to stop thinking about using the standard C++ library with managed types. The .Net BCL is really very good...
Hope this helps:
/*
the method demonstrates converting hexadecimal values,
which are broken into low and high bytes.
*/
int main(){
//character buffer
char buf[1];
buf[0]= 0x06; //buffer initialized to some hex value
buf[1]= 0xAE; //buffer initialized to some hex value
int number=0;
//number generated by binary shift of high byte and its OR with low byte
number = 0xFFFF&((buf[1]<<8)|buf[0]);
printf("%x",number); //this prints AE06
printf(“%d”,number); //this prints the integer equivalent
getch();
}