I'm not good in C, top of that I'm doing after very long, I need to do a very simple thing:
char code[]="aasd";
char *rmessage="";
strcat(rmessage,code[0]);
I simply want to concatenate the content of index 0 of array code to rmessage.
You need to ensure there is enough space in rmessage to store the result of the concatentation. You can use strncat to specify the number of characters to copy from a string:
char code[] = "aasd";
char rmessage[1024] = "";
strncat(rmessage, code, 1);
or, in this case, just assign the first character of rmessage:
rmessage[0] = code[0];
Not coding in C for long time.I think the syntax is just correct.
int sz=10; // sz = # number of chars you want to store + 1 , i assumed 9 characters will be stored at max.
char code[] = "aasd";
char *rmessage = malloc(sz*sizeof(char));
rmessage[0]=code[0];
rmessage[1]=NULL;
*Remember to deallocate the memory allocated to rmessage after your job is done.
free(rmessage);
Related
char input[5] = "12345";
printf("Convert to int valid %d", atoi(input));
printf("Convert to int invalid %d", atoi(input[1])); // program crash
Is there a solution to convert an char "slice" of an char string into an int?
Short description:
User inputs a string with values for example: 1, 2 3 4 ,5
Iam formating that string to 12345
With each number i want to continue to work with the index of an array.
If you mean "how to access a substring in the char [] array", you can use pointer arithmetic:
char input[6] = "12345";
printf("strtol(input + 1) = %d\n", strtol(input + 1, NULL, 10)); // Output = "2345";
Few things to note though:
Your array should be 6 elements long to hold the null terminator
atoi shouldn't be used at all; strtol is a better function for the task of converting a string to a signed integer; see here for more info.
Also, to convert a single character to an int:
if(isdigit(c))
{
c -= '0';
}
The relation that a textual representation of a digit is exactly '0' higher than the numeric value of that digit is guaranteed to hold for every character set supported by C.
To properly convert an arbitrary slice, you have to either make a copy or modify the string by inserting a \0 after the slice. The latter may not be an option, depending on where the string is stored.
To make a copy, allocate an array big enough to hold the slice and a \0. If you know the size of the slice at compile time, you can allocate on the stack:
char slice[2];
Otherwise, you'll have to allocate dynamically:
char *slice;
slice = malloc(2);
Stack allocated slices do not need to be deallocated, but dynamically allocated ones should be freed as soon as they are no longer needed:
free(slice);
Once you have the slice allocated, copy the portion of interest and terminate it with \0:
strncpy(slice, s + 1, 1);
slice[1] = '\0';
atoi(slice);
This technique will pretty much always work.
If your slice always ends with the string, you don't need to make a copy: you just need to pass a pointer to the start of the slice:
atoi(s + 1);
Modifying the string itself probably won't work, unless it's in writeable memory. If you're sure this is the case, you can do something like:
char tmp;
tmp = s[1];
s[1] = '\0';
atoi(s);
s[1] = tmp;
If you were sure but the memory wasn't writeable, your program will seg-fault.
For the special case where your slice is exactly one character long, you can use the fact that characters are numbers:
s[0] - '0'
Note that '0' !='\0' and that this won't work if your machine uses EBCDIC or similar.
I am new to C language. I need to concatenate char array and a char. In java we can use '+' operation but in C that is not allowed. Strcat and strcpy is also not working for me. How can I achieve this? My code is as follows
void myFunc(char prefix[], struct Tree *root) {
char tempPrefix[30];
strcpy(tempPrefix, prefix);
char label = root->label;
//I want to concat tempPrefix and label
My problem differs from concatenate char array in C as it concat char array with another but mine is a char array with a char
Rather simple really. The main concern is that tempPrefix should have enough space for the prefix + original character. Since C strings must be null terminated, your function shouldn't copy more than 28 characters of the prefix. It's 30(the size of the buffer) - 1 (the root label character) -1 (the terminating null character). Fortunately the standard library has the strncpy:
size_t const buffer_size = sizeof tempPrefix; // Only because tempPrefix is declared an array of characters in scope.
strncpy(tempPrefix, prefix, buffer_size - 3);
tempPrefix[buffer_size - 2] = root->label;
tempPrefix[buffer_size - 1] = '\0';
It's also worthwhile not to hard code the buffer size in the function calls, thus allowing you to increase its size with minimum changes.
If your buffer isn't an exact fit, some more legwork is needed. The approach is pretty much the same as before, but a call to strchr is required to complete the picture.
size_t const buffer_size = sizeof tempPrefix; // Only because tempPrefix is declared an array of characters in scope.
strncpy(tempPrefix, prefix, buffer_size - 3);
tempPrefix[buffer_size - 2] = tempPrefix[buffer_size - 1] = '\0';
*strchr(tempPrefix, '\0') = root->label;
We again copy no more than 28 characters. But explicitly pad the end with NUL bytes. Now, since strncpy fills the buffer with NUL bytes up to count in case the string being copied is shorter, in effect everything after the copied prefix is now \0. This is why I deference the result of strchr right away, it is guaranteed to point at a valid character. The first free space to be exact.
strXXX() family of functions mostly operate on strings (except the searching related ones), so you will not be able to use the library functions directly.
You can find out the position of the existing null-terminator, replace that with the char value you want to concatenate and add a null-terminator after that. However, you need to make sure you have got enough room left for the source to hold the concatenated string.
Something like this (not tested)
#define SIZ 30
//function
char tempPrefix[SIZ] = {0}; //initialize
strcpy(tempPrefix, prefix); //copy the string
char label = root->label; //take the char value
if (strlen(tempPrefix) < (SIZ -1)) //Check: Do we have room left?
{
int res = strchr(tempPrefix, '\0'); // find the current null
tempPrefix[res] = label; //replace with the value
tempPrefix[res + 1] = '\0'; //add a null to next index
}
I have parsed some date and time from GPS receiver. And need to convert them from string to int:
char year[4] = "2014";
char month[2] = "01";
char day[2] = "24";
char hour[2] ="12";
char minute[2] = "58";
char second[2] = "39";
GPS_current_year = atoi(year);
GPS_current_month = atoi(month);
GPS_current_day = atoi(day);
GPS_current_hour = atoi(hour);
GPS_current_minute = atoi(minute);
GPS_current_second = atoi(second);
After executing these the results are:
Somehow part of minutes string is converted when converting hour string. Same with minutes and seconds.
The strings are placed side by side in the memory.
If I change the sequence of defining strings then seconds may be added to years etc.
Questions:
What may cause this error?
Is there any way to avoid this error with using atoi?
I know that I can convert using a loop one char at a time. Just trying to find why is it not working.
Besides the missing quotes around the strings your char array's size should be defined to hold one more char the EOS (end of string a binary zero).
Since the memory representation would be e.g. "2014\0"
char year[4+1] = "2014";
Suggest not trying to define the string size as 4 or 5.
Let the compiler determine the string size.
char year[] = "2014";
In this case, the compiler will make year with a size of 5 initialized with '2', '0', '1', '4', '\0'.
OP's defining the size as 4 resulted in a size of 4-char array without a terminating '\0', which not being a string, create problems with atoi(year).
You forgot quotes in the strings:
char year[4] = "2014";
atoi() converts string to integer. But you are giving non string values to your string variables. Change your code to
char year[5] = "2014";
char month[3] = "01";
char day[3] = "24";
char hour[3] ="12";
char minute[3] = "58";
char second[3] = "39";
As simple as that. I'm on C++ btw. I've read the cplusplus.com's cstdlib library functions, but I can't find a simple function for this.
I know the length of the char, I only need to erase last three characters from it. I can use C++ string, but this is for handling files, which uses char*, and I don't want to do conversions from string to C char.
If you don't need to copy the string somewhere else and can change it
/* make sure strlen(name) >= 3 */
namelen = strlen(name); /* possibly you've saved the length previously */
name[namelen - 3] = 0;
If you need to copy it (because it's a string literal or you want to keep the original around)
/* make sure strlen(name) >= 3 */
namelen = strlen(name); /* possibly you've saved the length previously */
strncpy(copy, name, namelen - 3);
/* add a final null terminator */
copy[namelen - 3] = 0;
I think some of your post was lost in translation.
To truncate a string in C, you can simply insert a terminating null character in the desired position. All of the standard functions will then treat the string as having the new length.
#include <stdio.h>
#include <string.h>
int main(void)
{
char string[] = "one one two three five eight thirteen twenty-one";
printf("%s\n", string);
string[strlen(string) - 3] = '\0';
printf("%s\n", string);
return 0;
}
If you know the length of the string you can use pointer arithmetic to get a string with the last three characters:
const char* mystring = "abc123";
const int len = 6;
const char* substring = mystring + len - 3;
Please note that substring points to the same memory as mystring and is only valid as long as mystring is valid and left unchanged. The reason that this works is that a c string doesn't have any special markers at the beginning, only the NULL termination at the end.
I interpreted your question as wanting the last three characters, getting rid of the start, as opposed to how David Heffernan read it, one of us is obviously wrong.
bool TakeOutLastThreeChars(char* src, int len) {
if (len < 3) return false;
memset(src + len - 3, 0, 3);
return true;
}
I assume mutating the string memory is safe since you did say erase the last three characters. I'm just overwriting the last three characters with "NULL" or 0.
It might help to understand how C char* "strings" work:
You start reading them from the char that the char* points to until you hit a \0 char (or simply 0).
So if I have
char* str = "theFile.nam";
then str+3 represents the string File.nam.
But you want to remove the last three characters, so you want something like:
char str2[9];
strncpy (str2,str,8); // now str2 contains "theFile.#" where # is some character you don't know about
str2[8]='\0'; // now str2 contains "theFile.\0" and is a proper char* string.
I am making my first parallel application, but I am stuck with basics of C. I need to know, how to cast int to char and then how to append one char to another.
It you could help me please, i would be glad. Thank you.
You can use itoa function to convert the integer to a string.
You can use strcat function to append characters in a string at the end of another string.
If you want to convert a integer to a character, just do the following -
int a = 65;
char c = (char) a;
Note that since characters are smaller in size than integer, this casting may cause a loss of data. It's better to declare the character variable as unsigned in this case (though you may still lose data).
To do a light reading about type conversion, go here.
If you are still having trouble, comment on this answer.
Edit
Go here for a more suitable example of joining characters.
Also some more useful link is given below -
http://www.cplusplus.com/reference/clibrary/cstring/strncat/
http://www.cplusplus.com/reference/clibrary/cstring/strcat/
Second Edit
char msg[200];
int msgLength;
char rankString[200];
........... // Your message has arrived
msgLength = strlen(msg);
itoa(rank, rankString, 10); // I have assumed rank is the integer variable containing the rank id
strncat( msg, rankString, (200 - msgLength) ); // msg now contains previous msg + id
// You may loose some portion of id if message length + id string length is greater than 200
Third Edit
Go to this link. Here you will find an implementation of itoa. Use that instead.
Casting int to char is done simply by assigning with the type in parenthesis:
int i = 65535;
char c = (char)i;
Note: I thought that you might be losing data (as in the example), because the type sizes are different.
Appending characters to characters cannot be done (unless you mean arithmetics, then it's simple operators). You need to use strings, AKA arrays of characters, and <string.h> functions like strcat or sprintf.
Casting int to char involves losing data and the compiler will probably warn you.
Extracting a particular byte from an int sounds more reasonable and can be done like this:
number & 0x000000ff; /* first byte */
(number & 0x0000ff00) >> 8; /* second byte */
(number & 0x00ff0000) >> 16; /* third byte */
(number & 0xff000000) >> 24; /* fourth byte */
int myInt = 65;
char myChar = (char)myInt; // myChar should now be the letter A
char[20] myString = {0}; // make an empty string.
myString[0] = myChar;
myString[1] = myChar; // Now myString is "AA"
This should all be found in any intro to C book, or by some basic online searching.
int i = 100;
char c = (char)i;
There is no way to append one char to another. But you can create an array of chars and use it.