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';
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.
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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 doesnโt work because toupper is designed to capitalize chars and not strings, per the documentation. Iโve been making it overly complicated and itโs stressing me out. So to start, I created an โint N=strlen(string);โ, then created an array thatโs โchar upper[N];โ. Then I write a for loop written as(please forgive the terrible syntax Iโm about to write), โfor (int i = 0; i < N; i++) { toupper(upper[j]); }โ. What am I doing wrong?
suppose I have this line of code:
char myarr[10][0];
myarr[1][0] = "test";
printf("myarr: %c\n",myarr[1][0]);
Why does this give an error and how do I assign string to myarr?
You need a 2 dimensional character array to have an array of strings:
#include <stdio.h>
int main()
{
char strings[3][256];
scanf("%s %s %s", strings[0], strings[1], strings[2]);
printf("%s\n%s\n%s\n", strings[0], strings[1], strings[2]);
}
Use a 2-dimensional array char input[3][10];
or
an array of char pointers (like char *input[3];) which should be allocated memory dynamically before any value is saved at those locations.
First Case, take input values as scanf("%s", input[0]);, similarly for input[1] and input[2]. Remember you can store a string of max size 10 (including '\0' character) in each input[i].
In second case, get input the same way as above, but allocate memory to each pointer input[i] using malloc before. Here you have flexibility of size for each string.
If you don't want to change the strings, then you could simply do
const char *a[2];
a[0] = "blah";
a[1] = "hmm";
When you do it like this you will allocate an array of two pointers to const char. These pointers will then be set to the addresses of the static strings "blah" and "hmm".
If you do want to be able to change the actual string content, the you have to do something like
char a[2][14];
strcpy(a[0], "blah");
strcpy(a[1], "hmm");
This will allocate two consecutive arrays of 14 chars each, after which the content of the static strings will be copied into them.
There are several ways to create an array of strings in C. If all the strings are going to be the same length (or at least have the same maximum length), you simply declare a 2-d array of char and assign as necessary:
char strs[NUMBER_OF_STRINGS][STRING_LENGTH+1];
...
strcpy(strs[0], aString); // where aString is either an array or pointer to char
strcpy(strs[1], "foo");
You can add a list of initializers as well:
char strs[NUMBER_OF_STRINGS][STRING_LENGTH+1] = {"foo", "bar", "bletch", ...};
This assumes the size and number of strings in the initializer match up with your array dimensions. In this case, the contents of each string literal (which is itself a zero-terminated array of char) are copied to the memory allocated to strs. The problem with this approach is the possibility of internal fragmentation; if you have 99 strings that are 5 characters or less, but 1 string that's 20 characters long, 99 strings are going to have at least 15 unused characters; that's a waste of space.
Instead of using a 2-d array of char, you can store a 1-d array of pointers to char:
char *strs[NUMBER_OF_STRINGS];
Note that in this case, you've only allocated memory to hold the pointers to the strings; the memory for the strings themselves must be allocated elsewhere (either as static arrays or by using malloc() or calloc()). You can use the initializer list like the earlier example:
char *strs[NUMBER_OF_STRINGS] = {"foo", "bar", "bletch", ...};
Instead of copying the contents of the string constants, you're simply storing the pointers to them. Note that string constants may not be writable; you can reassign the pointer, like so:
strs[i] = "bar";
strs[i] = "foo";
But you may not be able to change the string's contents; i.e.,
strs[i] = "bar";
strcpy(strs[i], "foo");
may not be allowed.
You can use malloc() to dynamically allocate the buffer for each string and copy to that buffer:
strs[i] = malloc(strlen("foo") + 1);
strcpy(strs[i], "foo");
BTW,
char (*a[2])[14];
Declares a as a 2-element array of pointers to 14-element arrays of char.