It won't automatically convert (thank god). You'll have to use the method c_str() to get the C string version.
std::string str = "string";
const char *cstr = str.c_str();
.c_str() returns a const char *. If you want a non-const char *, use .data():
std::string str = "string";
char *cstr = str.data();
Some other options:
Copying the characters into a vector:
std::vector<char> cstr(str.c_str(), str.c_str() + str.size() + 1);Then
cstr.data()will give you the pointer.This version copies the terminating
\0. If you don't want it, remove+ 1or dostd::vector<char> cstr(str.begin(), str.end());.Copying into a manually allocated array: (should normally be avoided, as manual memory management is easy to get wrong)
std::string str = "string"; char *cstr = new char[str.size() + 1]; std::strcpy(cstr, str.c_str()); // do stuff delete [] cstr;
It won't automatically convert (thank god). You'll have to use the method c_str() to get the C string version.
std::string str = "string";
const char *cstr = str.c_str();
.c_str() returns a const char *. If you want a non-const char *, use .data():
std::string str = "string";
char *cstr = str.data();
Some other options:
Copying the characters into a vector:
std::vector<char> cstr(str.c_str(), str.c_str() + str.size() + 1);Then
cstr.data()will give you the pointer.This version copies the terminating
\0. If you don't want it, remove+ 1or dostd::vector<char> cstr(str.begin(), str.end());.Copying into a manually allocated array: (should normally be avoided, as manual memory management is easy to get wrong)
std::string str = "string"; char *cstr = new char[str.size() + 1]; std::strcpy(cstr, str.c_str()); // do stuff delete [] cstr;
More details here, and here but you can use
string str = "some string" ;
char *cstr = &str[0];
As of C++11, you can also use the str.data() member function, which returns char *
string str = "some string" ;
char *cstr = str.data();
To answer the question without reading too much else into it I would
Copychar str[2] = "\0"; /* gives {\0, \0} */
str[0] = fgetc(fp);
You could use the second line in a loop with whatever other string operations you want to keep using chars as strings.
You could do many of the given answers, but if you just want to do it to be able to use it with strcpy, then you could do the following:
Copy...
strcpy( ... , (char[2]) { (char) c, '\0' } );
...
The (char[2]) { (char) c, '\0' } part will temporarily generate null-terminated string out of a character c.
This way you could avoid creating new variables for something that you already have in your hands, provided that you'll only need that single-character string just once.
How to convert a string to character array in c (or) how to extract a single char form string? - Stack Overflow
String to char* conversion
How do I convert a string to an array of chars in C?
Simplest way to convert characters into a string? (C)
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In C, a string is actually stored as an array of characters, so the 'string pointer' is pointing to the first character. For instance,
char myString[] = "This is some text";
You can access any character as a simple char by using myString as an array, thus:
char myChar = myString[6];
printf("%c\n", myChar); // Prints s
Hope this helps! David
In C, there's no (real, distinct type of) strings. Every C "string" is an array of chars, zero terminated.
Therefore, to extract a character c at index i from string your_string, just use
char c = your_string[i];
Index is base 0 (first character is your_string[0], second is your_string[1]...).
I know that a string is already technically an array of chars, but when I try to use toupper(string), it doesn’t work because toupper is designed to capitalize chars and not strings, per the documentation. I’ve been making it overly complicated and it’s stressing me out. So to start, I created an “int N=strlen(string);”, then created an array that’s “char upper[N];”. Then I write a for loop written as(please forgive the terrible syntax I’m about to write), “for (int i = 0; i < N; i++) { toupper(upper[j]); }”. What am I doing wrong?
Hello everyone!
In this simple example, how can I replace the new_key string contents with the uppercase of the key string? I can print it alright with the below code, but can't find a way to store the characters in the empty string - there are always either segmentation fault errors or initializer ones.
What would be the best way to make new_key[i] be the content of toupper(key[i])?
#include <stdio.h>
#include <cs50.h>
#include <string.h>
#include <stdlib.h>
#include <ctype.h>
int main(void)
{
string key = "abcde";
string new_key[] = "";
for (int i = 0, len = strlen(key); i < len; i++)
{
char c = key[i];
printf("%c", toupper(c));
}
}
I've run into a lot of pointer related stuff recently, since then, one thing came up to my mind: "why does char* represent a string?"
and after this unsolved question, which i treated like some kind of axiom, I've ran into a new one, char**, the way I'm dealing with it feels like the same as dealing with an array of strings, and now I'm really curious about it
So, what's happening?
EDIT: i know strings doesn't exist in C and are represented by an array of char
You're saying you have this:
char array[20]; char string[100];
array[0]='1';
array[1]='7';
array[2]='8';
array[3]='.';
array[4]='9';
And you'd like to have this:
string[0]= "178.9"; // where it was stored 178.9 ....in position [0]
You can't have that. A char holds 1 character. That's it. A "string" in C is an array of characters followed by a sentinel character (NULL terminator).
Now if you want to copy the first x characters out of array to string you can do that with memcpy():
memcpy(string, array, x);
string[x] = '\0';
Assuming array is a character array that does not end in \0, you will want to use strncpy:
char * strncpy(char * destination, const char * source, size_t num);
like so:
strncpy(string, array, 20);
string[20] = '\0'
Then string will be a null terminated C string, as desired.