How to assign value to the pointer in pointers to an array

Hi,

i have code like this poinetrs to 'char' array.

char name[10];

char (*p) [10];

now i want to show the entered value for array 'name' to *p .
tried but
error 'type incompatible with assignment operator'
with,

*p = name;
*p = 'd';
*p = name[1];

can you giude me regarding this pointer to char array.

thank you.

char* p[10]

Yeah, you syntax is very wrong. First, why did you make an array of pointers? Secondly, and the reason it's having problems, is you're not supposed to have any unary operations on the lValue. You don't need any for assigning arrays to pointers anyway, but if for instance you were assigning a regular data type, then:

Code:

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// *p = integer1 should be
p = &integer1;

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but as I said, in the case of arrays, you're assigning the address of the array by default so you don't need the address of operator.

Code:

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#include <stdio.h>

int main() {
    char foo[13] = "Hello World!";
    char *bar;
   
    bar = foo;
   
    printf("%s", bar);
}

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Quote:

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Originally Posted by SlyMaelstrom

... First, why did you make an array of pointers?:

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It's actually a pointer to an array of chars, unless you're refering to Roshanx's example, which is an array of pointers - to char.

amas, I don't fully get your question but by the "*p" syntax use, it looks like all you really need is a character pointer: char* p = name;


xeddiex.

Ah yes, I see. Well in your case, then, the only thing you would need to make it run is the address of operator.

Code:

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p = &name;

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Though, I've never seen that syntax. The example I cited earlier is much more common.

Quote:

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Originally Posted by SlyMaelstrom

Ah yes, I see. Well in your case, then, the only thing you would need to make it run is the address of operator.

Code:

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p = &name;

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Though, I've never seen that syntax. The example I cited earlier is much more common.

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Not really, just the array name without the operator is fine because arrays automatically expose their address. I think you said this yourself in your first post .. hmm ;)


xeddiex.

Quote:

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Originally Posted by xeddiex

Not really, just the array name without the operator is fine because arrays automatically expose their address. I think you said this yourself in your first post .. hmm ;)

xeddiex.

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Sly is correct. It is a matter of type.

Code:

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char foo[10];

foo  // yields a pointer to character - char*
&foo // yields a pointer to a character array [10]  - char (*) [10];

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As Sly said, the second syntax is very rare. There are very few uses for it.