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.
Related
I'm trying to use sprintf() to put a string "inside itself", so I can change it to have an integer prefix. I was testing this on a character array of length 12 with "Hello World" inside it already.
The basic premise is that I want a prefix that denotes the amount of words within a string. So I copy 11 characters into a character array of length 12.
Then I try to put the integer followed by the string itself by using "%i%s" in the function. To get past the integer (I don't just use myStr as the argument for %s), I make sure to use myStr + snprintf(NULL, 0, "%i", wordCount), which should be myStr + characters taken up by the integer.
The problem is that I'm having is that it eats the 'H' when I do this and prints "2ello World" instead of having the '2' right beside the "Hello World"
So far I've tried different options for getting "past the integer" in the string when I try to copy it inside itself, but nothing really seems to be the right case, as it either comes out as an empty string or just the integer prefix itself '222222222222' copied throughout the entire array.
int main() {
char myStr[12];
strcpy(myStr, "Hello World");//11 Characters in length
int wordCount = 2;
//Put the integer wordCount followed by the string myStr (past whatever amount of characters the integer would take up) inside of myStr
sprintf(myStr, "%i%s", wordCount, myStr + snprintf(NULL, 0, "%i", wordCount));
printf("\nChanged myStr '%s'\n", myStr);//Prints '2ello World'
return 0;
}
First, to insert a one-digit prefix into a string “Hello World”, you need a buffer of 13 characters—one for the prefix, eleven for the characters in “Hello World”, and one for the terminating null character.
Second, you should not pass a buffer to snprintf as both the output buffer and an input string. Its behavior is not defined by the C standard when objects passed to it overlap.
Below is a program that shows you how to insert a prefix by moving the string with memmove. This is largely tutorial, as it is not generally a good way to manipulate strings. For short strings, where space is not an issue, most programmers would simply print the desired string into a temporary buffer, avoiding overlap issues.
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
/* Insert a decimal numeral for Prefix into the beginning of String.
Length specifies the total number of bytes available at String.
*/
static void InsertPrefix(char *String, size_t Length, int Prefix)
{
// Find out how many characters the numeral needs.
int CharactersNeeded = snprintf(NULL, 0, "%i", Prefix);
// Find the current string length.
size_t Current = strlen(String);
/* Test whether there is enough space for the prefix, the current string,
and the terminating null character.
*/
if (Length < CharactersNeeded + Current + 1)
{
fprintf(stderr,
"Error, not enough space in string to insert prefix.\n");
exit(EXIT_FAILURE);
}
// Move the string to make room for the prefix.
memmove(String + CharactersNeeded, String, Current + 1);
/* Remember the first character, because snprintf will overwrite it with a
null character.
*/
char Temporary = String[0];
// Write the prefix, including a terminating null character.
snprintf(String, CharactersNeeded + 1, "%i", Prefix);
// Restore the first character of the original string.
String[CharactersNeeded] = Temporary;
}
int main(void)
{
char MyString[13] = "Hello World";
InsertPrefix(MyString, sizeof MyString, 2);
printf("Result = \"%s\".\n", MyString);
}
The best way to deal with this is to create another buffer to output to, and then if you really need to copy back to the source string then copy it back once the new copy is created.
There are other ways to "optimise" this if you really needed to, like putting your source string into the middle of the buffer so you can append and change the string pointer for the source (not recommended, unless you are running on an embedded target with limited RAM and the buffer is huge). Remember code is for people to read so best to keep it clean and easy to read.
#define MAX_BUFFER_SIZE 128
int main() {
char srcString[MAX_BUFFER_SIZE];
char destString[MAX_BUFFER_SIZE];
strncpy(srcString, "Hello World", MAX_BUFFER_SIZE);
int wordCount = 2;
snprintf(destString, MAX_BUFFER_SIZE, "%i%s", wordCount, srcString);
printf("Changed string '%s'\n", destString);
// Or if you really want the string put back into srcString then:
strncpy(srcString, destString, MAX_BUFFER_SIZE);
printf("Changed string in source '%s'\n", srcString);
return 0;
}
Notes:
To be safer protecting overflows in memory you should use strncpy and snprintf.
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
}
Let's say that I have the following character array. I need to know if there is a way to get the length of the second string in the array. I know I could use strlen, but this gives me the length of the first string.
#include <stdio.h>
int main()
{
char greeting[20]={'P','H','O','N','E','\0','6','0','1','7','1','2','1','2','1','6','\0'};
int count;
count=strlen(greeting); // this returns the length of first string which is 5
}
A valid string in C is null terminated
Having 2 null characters in the char array might cause lot of issues especially while using inbuilt functions.
If you really want to do this
strlen(greeting + 1 + strlen(greeting));
strlen() works on a null terminated strings and if you pass just greeting to it then the length will be calculated till the first null is encountered and length will be returned so now we need length from the character after first null so you should calculate as shown above.
Simple, generic, readable way:
size_t s_length [SUBSTRINGS];
const char* strp = greeting;
for(size_t i=0; i<SUBSTRINGS; i++)
{
s_length[i] = strlen(strp);
strp += s_length[i] + 1; // +1 to skip null termination
}
I'm using the code below to add some "0" chars into my string, but it seems there is a problem and the program will crash. Everything seems logic but I do not know where is the problem?
#include <stdlib.h>
#include <string.h>
int main()
{
char *Ten; int i=0; Ten = malloc(12);
Ten="1";
for (i=0;i<10;i++)
strcat(Ten,"0");
printf("%s",Ten);
return 0;
}
You declare Ten as a pointer to a string literal. However, you cannot rely on being able to modify a string literal, and thus the program crashes.
To fix this, you can declare Ten as an array instead:
int main()
{
char Ten[12]="1"; int i=0;
for (i=0;i<10;i++)
strcat(Ten,"0");
printf("%s",Ten);
return 0;
}
Note that you need 12 bytes; 11 for the characters and one for the terminating NUL character.
Ten is a string literal and you cannot modify it. Try with array instead
char Ten[12] = "1";
for (i=0;i<10;i++)
strcat(Ten,"0");
printf("%s",Ten);
notice that I created an array of 12 characters, because there should be room for a termination '\0'.
You actually don't need strcat here, it's just do this
char Ten = malloc(12);
if (Ten != NULL)
{
Ten[0] = '1';
for (i = 1 ; i < 11 ; i++)
Ten[i] = '0';
Ten[11] = '\0';
/* Use Ten here, for example printf it. */
printf("%s",Ten);
/* You should release memory. */
free(Ten);
}
or
char Ten = malloc(12);
if (Ten != NULL)
{
Ten[0] = '1';
memset(Ten + 1, '0', 10);
Ten[11] = '\0';
/* Use Ten here, for example printf it. */
printf("%s",Ten);
/* You should release memory. */
free(Ten);
}
To quote from strcat manual on linux:
The strcat() function appends the src string to the dest string,
overwriting the terminating null byte ('\0') at the end of dest, and
then adds a terminating null byte. The strings may not overlap, and
the dest string must have enough space for the result. If dest is not
large enough, program behavior is unpredictable; buffer overruns are
a favorite avenue for attacking secure programs.
Your Ten array is only long enough to store original literal. You need to preallocate memory as long as final desired string.
String literals might be stored in read only section of memory. Any attempt to modify such a literal causes undefined behavior.
To concatenate two strings, the destination must have enough space allocated for the characters to be added and space for '\0'. Change the declaration of Ten to
char Ten[12] = "1";
and it will work.
I am processing an input string, which consists of a process name, followed by an arbitrary amount of arguments.
I need the process name , along with all of the arguments, in one string.
I thought I could use strcat in a loop, so that it cycles through all of the args and each time appends the arg to the string, but I am having problems with getting a string that in empty to begin the loop.
Can anyone help me out with some basic code?
Thanks.
EDIT:
I'm posting my code for clarity. Mike's post is closest to what I have now:
char * temp;
strcpy(temp,"");
for (i = 4; i < argc-1; i++) // last arg is null, so we need argc-1
{
strcat(temp,argv[i]);
strcat(temp," ");
}
ignore the 4 in my for loop for the moment (magic number, i know.)
I am getting a segfault with this code. Is it because of my string assignment? I assume that is the case and hence I asked the question of how i could combine the strings.
Let's say your input strings are in an array of char pointers, suggestively called argv, of length argc.
We first need to determine how much space is needed for the output:
int length = 0;
for (int i = 0; i < argc; ++i)
length += strlen(argv[i]);
Then we allocate it, adding an extra char for the '\0' terminator:
char *output = (char*)malloc(length + 1);
Finally, the concatenation:
char *dest = output;
for (int i = 0; i < argc; ++i) {
char *src = argv[i];
while (*src)
*dest++ = *src++;
}
*dest = '\0';
Note that I don't use strcat here. Reason is that this sets us up for a Schlemiel the Painter's algorithm: for each iteration, the entire output string would be scanned to find its end, resulting in quadratic running time.
Don't forget to free the output string when you're done:
free(output);
I'm a bit tired so I may be overlooking something here. A better solution, using standard library functions, is welcome. It would be convenient if strcat returned a pointer to the terminator byte in dest, but alas.
You want an empty C string? Is this what you are looking for: char p[] = "";?
UPDATE
After you posted some code it is clear that you have forgotten to allocate the buffer temp. Simply run around the arguments first, counting up the length required (using strlen), and then allocate temp. Don't forget space for the zero terminator!
You could provide the "arbitrary amount of arguments" as one argument, ie an array/list, then do this pseudocode:
str = "";
i = 0;
while i < length of input
{
str = strcat ( str , input[i]);
i++;
}
#include<stdio.h>
#include<stdarg.h>
int main(int argc, char** argv) {
// the main parameters are the same situation you described
// calling this program with main.exe asdf 123 fdsa, the program prints out: asdf123fdsa
int lengths[argc];
int sum =0;
int i;
for(i=1; i<argc; i++) { // starting with 1 because first arg is program-path
int len = strlen(argv[i]);
lengths[i] = len;
sum+=len;
}
char* all = malloc(sum+1);
char* writer = all;
for(i=1; i<argc; i++) {
memcpy(writer, argv[i], lengths[i]);
writer+=lengths[i];
}
*writer = '\0';
printf("%s\n", all);
system("pause");
return 0;
}
A string in C is represented by an array of characters that is terminated by an "null" character, '\0' which has the value 0. This lets all string functions know where the end of a string is. Here's an exploration of different ways to declare an empty string, and what they mean.
The usual way of getting an empty string would be
char* emptyString = "";
However, emptyString now points to a string literal, which cannot be modified. If you then want to concatenate to an empty string in your loop, you have to declare it as an array when you initialize.
char buffer[] = "";
This gives you an array of size one. I.e. buffer[0] is 0. But you want an array to concatenate to- it has to be large enough to accomodate the strings. So if you have a string buffer of certain size, you can initialize it to be empty like so:
char buffer[256] = "";
The string at buffer is now "an empty string". What it contains, is buffer[0] is 0 and the rest of the entries of the buffer might be garbage, but those will be filled once you concatenate your other strings.
Unfortunately, the problem with C, is you can never have an "infinite" string, where you are safe to keep concatenating to, you have to know it's definite size from the start. If your array of arguments are also strings, you can find their length using strlen. This gives you the length of a string, without the null character. Once you know the lengths of all your sub-strings, you will now know how long your final buffer will be.
int totalSize; // assume this holds the size of your final concatenated string
// Allocate enough memory for the string. the +1 is for the null terminator
char* buffer = malloc(sizeof(char) * (totalSize + 1));
buffer[0] = 0; // The string is now seen as empty.
After this, you are free to concatenate your strings using strcat.