0 
#include <stdio.h>
int main(void){
    char c1[], c2[];
    c1 = "hello";
    c2 = "world";
    printf("%s %s", c1, c2);
    return 0;
}

What exactly makes it that I cannot use 'char c1[], c2[]'? I have some knowledge in Java and I find C syntax to be rather familiar with it but clearly somethings do not work.

Is there also any reason why char is declared as char variable[] instead of char[] variable? It seems to make more sense like this (Java notation)

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7
  • I never use c for long time but if I remember array must defined length Commented May 4, 2016 at 4:05
  • you write c[2] = 'x'; and not [2]c = 'x';, so you write char c[3]; and not char[3] c; . Commented May 4, 2016 at 4:11
  • char c1[] - so how many characters do you expect c1 to hold? In C, this makes c1 an array, not a reference to an array like it does in Java. Commented May 4, 2016 at 4:22
  • 1
    Just forget about Java and study a beginner-level book in C. Commented May 4, 2016 at 6:22
  • regarding this kind of expression: char c1[], An array in C, must be either declared with a specific length or initialized with a char string. I.E. either: char c1[20]; or char c1[] = "hello ";. Strings can only be copied char by char or using a function like: strcpy() Only at variable initialization can a char array be set using =. Commented May 7, 2016 at 1:56

2 Answers 2

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6

This:

char c1[]

Means "An array of characters with unknown length." You can't create arrays of unknown length, though you can take them as parameters to a function, as in:

void foo(char c1[])

You can also create them if the length is known implicitly, as in:

char c1[] = "hello"; // size is 6, including null terminator

And as for this:

Is there also any reason why char is declared as char variable[] instead of char[] variable?

It has always been like this. I'm sure C programmers feel the Java way doesn't make sense too. Decisions made in language syntax 40 years ago are not usefully debated today.

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4 Comments

But why is it in the case that char c1[] = "hello"; it is able to figure it out on its own whilst if you have more than one character array it suddenly needs the direct length? What if I wanted to input a giant string of text that I didnt know the length for?
Specifically, it says "make me an array of unknown length on the stack". Which makes no sense. In Java it means "make me a pointer to an array on the stack", this is because of Java's concept of 'value type' and 'reference type'.
I dont understand... in the code above if I removed c2 altogether the code works fine... yet I didnt declare the size of the array. Why does this work?
Technically OP's array would gain an implicit size in the first assignment. Hypothetically the compiler could attempt to find a first assignment, so I think a little bit more needs to be explained. "hello" is known at compile time after all.
1

That code precisely speaking is not declarations, it's definitions. And in C, you cannot define arrays leaving the length empty, you can either specify it explicitly:

char c1[6] = "hello";

or you can do it implicitly:

char c1[] = "hello";

If the length of the array is only known in running time, use allocated arrays:

size_t n = <calculate the size needed>;
char *c1 = malloc(n);

But for declarations, you can do external declarations like this:

extern char c1[];
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13 Comments

So let us say id want to declare a string array that contains well over 1000 characters, how would I know what to input?
@Wjk just do char c[1000];?
@Wjk If the size is not known at developing time and it is calculated in runtime, you need to use dynamic allocation: size_t n = get_size(); char *c1 = malloc(n);
Or you could simply do char c1[n].
@Lundin Yes, that will also do.
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