There are two way of working with array of characters (strings) in C. They are as follows:
char a[ROW][COL];
char *b[ROW];
Pictorial representation is available as an inline comment in the code.
Based on how you want to represent the array of characters (strings), you can define pointer to that as follows
char (*ptr1)[COL] = a;
char **ptr2 = b;
They are fundamentally different types (in a subtle way) and so the pointers to them is also slightly different.
The following example demonstrates the different ways of working with strings in C and I hope it helps you in better understanding of array of characters (strings) in C.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define ROW 5
#define COL 10
int main(void)
{
int i, j;
char a[ROW][COL] = {"string1", "string2", "string3", "string4", "string5"};
char *b[ROW];
/*
a[][]
0 1 2 3 4 5 6 7 8 9
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 1 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 2 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 3 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 4 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 5 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
*/
/* Now, lets work on b */
for (i=0 ; i<5; i++) {
if ((b[i] = malloc(sizeof(char) * COL)) == NULL) {
printf("unable to allocate memory \n");
return -1;
}
}
strcpy(b[0], "string1");
strcpy(b[1], "string2");
strcpy(b[2], "string3");
strcpy(b[3], "string4");
strcpy(b[4], "string5");
/*
b[] 0 1 2 3 4 5 6 7 8 9
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 1 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 2 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 3 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 4 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 5 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
*/
char (*ptr1)[COL] = a;
printf("Contents of first array \n");
for (i=0; i<ROW; i++)
printf("%s \n", *ptr1++);
char **ptr2 = b;
printf("Contents of second array \n");
for (i=0; i<ROW; i++)
printf("%s \n", ptr2[i]);
/* b should be free'd */
for (i=0 ; i<5; i++)
free(b[i]);
return 0;
}
Answer from Sangeeth Saravanaraj on Stack OverflowThere are two way of working with array of characters (strings) in C. They are as follows:
char a[ROW][COL];
char *b[ROW];
Pictorial representation is available as an inline comment in the code.
Based on how you want to represent the array of characters (strings), you can define pointer to that as follows
char (*ptr1)[COL] = a;
char **ptr2 = b;
They are fundamentally different types (in a subtle way) and so the pointers to them is also slightly different.
The following example demonstrates the different ways of working with strings in C and I hope it helps you in better understanding of array of characters (strings) in C.
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define ROW 5
#define COL 10
int main(void)
{
int i, j;
char a[ROW][COL] = {"string1", "string2", "string3", "string4", "string5"};
char *b[ROW];
/*
a[][]
0 1 2 3 4 5 6 7 8 9
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 1 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 2 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 3 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 4 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
| s | t | r | i | n | g | 5 | '\0' | | |
+---+---+---+---+---+---+---+------+---+---+
*/
/* Now, lets work on b */
for (i=0 ; i<5; i++) {
if ((b[i] = malloc(sizeof(char) * COL)) == NULL) {
printf("unable to allocate memory \n");
return -1;
}
}
strcpy(b[0], "string1");
strcpy(b[1], "string2");
strcpy(b[2], "string3");
strcpy(b[3], "string4");
strcpy(b[4], "string5");
/*
b[] 0 1 2 3 4 5 6 7 8 9
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 1 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 2 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 3 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 4 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
| --|------->| s | t | r | i | n | g | 5 | '\0' | | |
+--------+ +---+---+---+---+---+---+---+------+---+---+
*/
char (*ptr1)[COL] = a;
printf("Contents of first array \n");
for (i=0; i<ROW; i++)
printf("%s \n", *ptr1++);
char **ptr2 = b;
printf("Contents of second array \n");
for (i=0; i<ROW; i++)
printf("%s \n", ptr2[i]);
/* b should be free'd */
for (i=0 ; i<5; i++)
free(b[i]);
return 0;
}
What would be the correct way to solve this problem?
Well, the correct way would be to use a library specifically designed for dealing with multilanguage interfaces - for instance gettext.
Another way, though patchier, would be to use a hash table (also known as "dictionary" or "hash map" or "associative map" in other languages/technologies): Looking for a good hash table implementation in C
It's probably not the answer you were looking for, but you've asked the wrong question to the right problem.
Videos
The issue is that you are not allocating any space for those names. You need to initialize each element in the array if you intend to use it with scanf.
char* names[6];
for( int i = 0; i < 6; ++i )
names[i] = malloc( 256 * sizeof *names[i] ); // or some other max value
scanf( "%s", names[1] );
Otherwise those pointers will be pointing anywhere in your memory, and attempting to read/write those locations will eventually result in a segmentation fault.
In your code names is an array of 6 pointers to char. Now each of these pointers can store the starting point (the address of the first character) of a new string. This means you can store the starting addresses of 6 different strings in your names variable.
But when you use a loop to initialize each of these strings, you need to inform the machine HOW long each string might be, so that it can allocate a continuous block of addresses whose first address can then be stored in your pointer to refer to your string. Thus, you must allocate a certain size you think should be sufficient to store your string (eg: 256 bytes, 1 byte being 1 character). In the absence of this, the machine doesn't know where to store all the bytes of your string and throws a segmentation fault due to illegal memory access.
Thus to do this, each of your 6 pointers must be allocated some space to store a string. This will be done in your loop using malloc(). Based on @K-ballo's code:
char* names[6];
int max_length = 256; // The maximum length you expect
for( int i = 0; i < 6; ++i )
names[i] = malloc( max_length * sizeof(char) ); // allocates max_length number of bytes
scanf( "%s", names[1] );
So now you basically have a 6 different blocks of max_length continuous char addresses that are each referred to by names[i]. When you do the scanf() it reads the bytes from standard input and puts then into these allocated bytes in memory referred to by names[1].
I had a difficult time at the start understanding all this, so just thought an elaborate explanation would help. :)
I've tried everything! I've never been this stuck!
static char file[ROWS][LENGTH];
void loadFile(FILE* file, char** buffer){
// load the file
}
void saveFile(FILEI* file, char** buffer){
// save the file
}
int main(){
FILE* theFile = fopen("the path.txt", "r");
loadFile(theFile, file);
fclose(theFile);
return 0;
}
I've tried using char** buffer, char* buffer[], char* buffer[][] in the function prototype
and I've tried passing file, &file[0], file[], &file into the function call.
but I get "warning: passing argument 2 of 'loadFile' from incompatible pointer type"
I need a way to make file[ROWS][LENGTH]; work. How do I do this?
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.
Given that the code is really simple, I see mostly coding style issues with it.
Instead of this:
char *names[] = { "John", "Mona", "Lisa", "Frank" };
I would prefer either of these writing styles:
char *names[] = { "John", "Mona", "Lisa", "Frank" };
// or
char *names[] = {
"John",
"Mona",
"Lisa",
"Frank"
};
The pNames variable is pointless. You could just use names.
Instead of the while loop, a for loop would be more natural.
This maybe a matter of taste,
but I don't think the Hungarian notation like *pArr is great.
And in any case you are using this pointer to step over character by character,
so "Arr" is hardly a good name.
I'd for go for pos instead. Or even just p.
You should declare variables in the smallest scope where they are used.
For example *pos would be best declared inside the for loop.
In C99 and above, the loop variable can be declared directly in the for statement.
The last return statement is unnecessary.
The compiler will insert it automatically and make the main method return with 0 (= success).
Putting it together:
int main(int argc, char *argv[])
{
char *names[] = { "John", "Mona", "Lisa", "Frank" };
for (int i = 0; i < 4; ++i) {
char *pos = names[i];
while (*pos != '\0') {
printf("%c\n", *(pos++));
}
printf("\n");
}
}
Actually it would be more interesting to use argc and argv for something:
int main(int argc, char *argv[])
{
for (int i = 1; i < argc; ++i) {
char *pos = argv[i];
while (*pos != '\0') {
printf("%c\n", *(pos++));
}
printf("\n");
}
}
Other points not already mentioned:
Use const where possible
It's better to use const where possible so that it's clear when you are intending to modify things and clear when you're not. This helps the compiler help you find bugs early.
Avoid magic numbers
It's better to avoid having numbers like 4 in the program. If you want to change things later, you may well be left wondering "why 4? what does that mean?" Better would be to assign a constant with a meaningful name.
Use putchar rather than printf for single characters
The printf function is very useful, but putchar is often better suited for single-character output. The reason is that printf has to, at runtime, parse the format string, apply the arguments and then emit the results, while putchar only has to pass a character directly to the output. This particular code isn't exactly performance-critical but it's useful to acquire good coding habits from the beginning.
Use C idioms
It's more idiomatic C to have a loop like this for the characters:
while(*pArr) { /* ... */ }
rather than this:
while(*pArr != '\0') { /* ... */ }
Consider using a "sentinel" value for the end of a list
There are two common ways to iterate through a list in C. One is as you already have it, when you know how many values are in the list. The other way is to use a "sentinel" value -- some unique value that tells the program "this is the end of the list." For a list of names such as this program has, a rational choice for a sentinel value would be NULL.
Omit the return 0 from the end of main
Uniquely for main, if the program gets to the end of the function, it automatically does the equivalent of return 0 at the end, so you can (and should) omit it.
Result
Using all of these suggestions, one possible rewrite might look like this:
#include <stdio.h>
int main()
{
const char *names[] = { "John", "Mona", "Lisa", "Frank", NULL };
for (int i=0; names[i]; ++i) {
const char *ch = names[i];
while(*ch) {
putchar(*ch++);
putchar('\n');
}
putchar('\n');
}
}
So i am learning command lines arguments and just came cross char *argv[]. What does this actually do, I understand that this makes every element in the array a pointer to char, but i can't get around as to how all of this is happening. How does it treat every other element as another string? How come because essentialy as of my understanding rn, a simple char would treat as a single contiguous block of memory, how come turning this pointer to another pointer of char point to individual elements of string?