I have the the ascii value of the integer number stored in string variable in arduino sketch and I want to convert it to the integer number. How do I do this?
String a = "59"; // or, 0x32, ASCII value of integer number 2
const char * s = &a[0];
int num = atoi(s);
I expected the num to be 2 (the number corresponding the ascii 59)
But, when I print 'num' in serial monitor, I am getting it to be 59 (Not 2).
You need to cast your integer to char, when sending it. As in:
int x = atoi("50");
Serial.write((char)x);
Related
for (int x= 7; 0<=x; x--) {
size_t x_val = ((1<<4)-1) & io>>x*4;
printf("%lX", x_val);
Trying to print a hexadecimal number here after converting it from integer. While the conversion is successful, the output is FFFFFFCD instead of the desired output FFCD. How can I limit maximum 4 characters to be printed?
to print a four character hexadecimal
Limit value to the [0...0xFFFF] range.
Print 4 digits padded with zeros with the correct specifier.
Example
unsigned masked_value = io & 0xFFFF;
printf("%04X\n", masked_value);
Try this one,
You can use hh to tell printf that the argument is an unsigned char. Use 0 to get zero padding and 4 to set the width to 4. x or X for lower/uppercase hex characters.
uint8_t a = 0x0a;
printf("%02hhX", a); // Prints "0A"
printf("0x%02hhx", a);
I am having issues with converting character variables into integer variables. This is my code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int main()
{
char string[] = "A2";
char letter = string[0];
char number = string[1];
char numbers[] = "12345678";
char letters[] = "ABCDEFGH";
int row;
int column;
for(int i = 0; i < 8; i++){
if(number == numbers[i]){
row = number;
}
}
}
When I try to convert the variable row into the integer value of the variable number, instead of 2 I get 50. The goal so far is to convert the variable row into the accurate value of the character variable number, which in this case is 2. I'm a little confused as to why the variable row is 50 and not 2. Can any one explain to me why it is not converting accurately?
'2' != 2. The '2' character, in ASCII, is 50 in decimal (0x32 in hex). See http://www.asciitable.com/
If you're sure they're really numbers you can just use (numbers[i] - '0') to get the value you're looking for.
2 in your case is a character, and that character's value is 50 because that's the decimal version of the byte value that represents the character 2 in ASCII. Remember, c is very low level and characters are essentially the same thing as any other value: a sequence of bytes. Just like letters are represented as bytes, so are the character representation of their value in our base 10 system. It might seem that 2 should have been represented with the value 2, but it wasn't.
If you use the atoi function, it will look at the string and compute the decimal value represented by the characters in your string.
However, if you're only converting one character to the decimal value it represents , you can take a short cut. subtract the digit from the value of '0'. Though the digits are not represented by the base 10 value they have for us humans, they are ordered sequentially in the ASCII code. And since in C the characters are simply byte values, the difference between a numeric character 0-9 and 0 is the value of the character.
char c = '2';
int i = c - '0';
If you understand why that would work, you get what I'm saying.
I'm looking to convert a input variable of this type:
char var[] = "9876543210000"
to a hex equivalent char
char hex[] = "08FB8FD98210"
I can perform this by e.g. the following code::
long long int freq;
char hex[12];
freq = strtoll(var,NULL,10);
sprintf(hex, "%llX",freq);
However I'm doing this on a avr microcontroller, thus strtoll is not available only strtol (avr-libgcc). So I'm restricted to 32 bits integer which is not enough. Any ideas?
Best regards
Simon
Yes.... this method works fine only with positive number, so if you have a minus sign, just save it before doing next. Just divide the string number in two halves, lets say that you select six digit each, to get two decimal numbers: u_int32_t left_part; and u_int32_t right_part; with the two halves of your number.... you can construct your 64 bit number as follows:
u_int64_t number = (u_int64_t) left_part * 1000000 + right_part;
If you have the same problem on the printing side, that is, you cannot print but 32 bit hex numbers, you can just get left_part = number >> 32; and right_part = number & 0xffffffff; and finally print both with:
if (left_part) printf("%x%08x", left_part, right_part);
else printf("%x", right_part);
the test makes result not to be forced to 8 digits when it is less than 0x100000000.
It looks like you might have to parse the input one digit at a time, and save the result into a uint64_t variable.
Initialize the uint64_t result variable to 0. In a loop, multiply the result by 10 and add the next digit converted to an int.
Now to print the number out in hex, you can use sprintf() twice. First print result >> 32 as a long unsigned int, followed by (long unsigned int)result at &hex[4](or 6 or 8 or wherever) to pick up the remaining 32 bits.
You will need to specify the format correctly to get the characters in the array in the correct places. Perhaps, just pad it with 0s? Don't forget about room for the trailing null character.
Change this:
freq = strtoll(var,NULL,10);
To this:
sscanf(var,"%lld",&freq);
So I have a string like this: "MULTR12"
I want to get '1' and '2' as two seperate integers. How do I go about doing that? Before, I had it simply as:
char *string = "MULTR12";
val = string[5];
But I get a really weird value for val instead of 1, like I want. Can anyone help? Thanks!
This is how you convert a char to int ..
int x = string[5] - '0';
Here's some explanation.. Every character is represented as an ASCII character in memory, A will be 65 and simliar. This also applies to numbers, so 0 in ASCII is 48, 1 is 49 etc.
Now, when we subtract 0 from any number's ASCII representation. Let's say the number is 5, this is what we are actually doing..
int x = 53 - 48
which gives us 5 as an integer. In other words, we are calculating the displacement of that numbers ASCII representation from 0's ASCII representation.
A bit hackish but try
val = string[5] - '0';
Shamelessly stolen from here
What you are trying to do above, is basically take the ASCII representation of the char '1' and convert it to an int, yielding 49. (probably not what you expected)
You can do something like this:
int num = string[ 5 ] - '0';
Actually you are getting a weird value because you are trying to change the value of a string that is fixed.
First take a new array as char str[]="Multr12";
Now you can access str[5]...try this
I have a little (big, dumb?) question about int and chars in C. I rememeber from my studies that "chars are little integers and viceversa," and that's okay to me. If I need to use small numbers, the best way is to use a char type.
But in a code like this:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char *argv[]) {
int i= atoi(argv[1]);
printf("%d -> %c\n",i,i);
return 0;
}
I can use as argument every number I want. So with 0-127 I obtain the expected results (the standard ASCII table) but even with bigger or negative numbers it seems to work...
Here is some example:
-181 -> K
-182 -> J
300 -> ,
301 -> -
Why? It seems to me that it's cycling around the ascii table, but I don't understand how.
When you pass an int corresponding to the "%c" conversion specifier, the int is converted to an unsigned char and then written.
The values you pass are being converted to different values when they are outside the range of an unsigned (0 to UCHAR_MAX). The system you are working on probably has UCHAR_MAX == 255.
When converting an int to an unsigned char:
If the value is larger than
UCHAR_MAX, (UCHAR_MAX+1) is
subtracted from the value as many
times as needed to bring it into the
range 0 to UCHAR_MAX.
Likewise, if the
value is less than zero, (UCHAR_MAX+1)
is added to the value as many times
as needed to bring it into the range
0 to UCHAR_MAX.
Therefore:
(unsigned char)-181 == (-181 + (255+1)) == 75 == 'K'
(unsigned char)-182 == (-182 + (255+1)) == 74 == 'J'
(unsigned char)300 == (300 - (255+1)) == 44 == ','
(unsigned char)301 == (301 - (255+1)) == 45 == '-'
The %c format parameter interprets the corresponding value as a character, not as an integer. However, when you lie to printf and pass an int in what you tell it is a char, its internal manipulation of the value (to get a char back, as a char is normally passed as an int anyway, with varargs) happens to yield the values you see.
My guess is that %c takes the first byte of the value provided and formats that as a character. On a little-endian system such as a PC running Windows, that byte would represent the least-significant byte of any value passed in, so consecutive numbers would always be shown as different characters.
You told it the number is a char, so it's going to try every way it can to treat it as one, despite being far too big.
Looking at what you got, since J and K are in that order, I'd say it's using the integer % 128 to make sure it fits in the legal range.
Edit: Please disregard this "answer".
Because you are on a little-endian machine :)
Serously, this is an undefined behavior. Try changing the code to printf("%d -> %c, %c\n",i,i,'4'); and see what happens then...
When we use the %c in printf statement, it can access only the first byte of the integer.
Hence anything greater than 256 is treated as n % 256.
For example
i/p = 321 yields op=A
What atoi does is converting the string to numerical values, so that "1234" gets 1234 and not just a sequence of the ordinal numbers of the string.
Example:
char *x = "1234"; // x[0] = 49, x[1] = 50, x[2] = 51, x[3] = 52 (see the ASCII table)
int y = atoi(x); // y = 1234
int z = (int)x[0]; // z = 49 which is not what one would want