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if an array char is already a pointer, why declare a pointer to the array?
And what’s the difference between char exemplo[10] and char *exemplo[10]?
2 answers
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An array char (or other type) IS NOT a pointer.
An array is an array; a pointer is a pointer (see Section 6 of c-Faq).
When an array is used as value, it is converted to a pointer to its first element.
The difference between char exemplo[10] and char *exemplo[10] is that the first declares a 10-character array and the second declares a 10-pointer array.
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+1 by "A char array (or other type) IS NOT a pointer". I have seen several responses indicating: in the C language arrays do not actually exist, or that they are the same as pointers.
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You nay is declaring a pointer to array. Is declaring a pointer to char. Ai contrary to popular belief they are separate things.
In the first example there is a array with 10 chars. In the second there is a array with 10 hands for char.
The pointer to char is practically a synonym for string. C language does not have the type string, but this is as close to a.
Obviously the pointer needs to point to an area of memory that has a sequence of chars.
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A literal between quotation marks, in C, has an array of
Ncharacters in whichNis sufficient for the characters of the literal and the terminator."exemplo"has kindchar [8]. -
@pmg is true, it decays to a
const char *just. I fixed it. -
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When used as value, the array is converted to a pointer to its first element. In the case of literal string this conversion is to the type
char *(nayconst char *) although the result is read-only. To avoid surprises many people add theconstspecifically.
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