Passing address or pointer as argument to function

Code:

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char_t buffer[10];

somefunction(&buffer);

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Code:

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char_t* pbuffer;
pbuffer = (char_t*)malloc(10);

somefunction(pbuffer);

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somefunction will be filling up the buffer with data

Is there any difference between the above 2? From what I understand there is no difference since both are passing a pointer to the function. Am I right?

Yes, the first one is wrong. Arrays decay to pointers when passed to functions so you can drop the address of operator, e.g.

Code:

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char_t buffer[10];
somefunction(buffer);

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oh so if we want to pass a char array into a function, we just have to pass in the name of the array? like in your example?

Quote:

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Originally Posted by Edelweiss

like in your example?

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No, we just like to make up stuff here. :rolleyes:


Quzah.

Yes, when you pass the name of the array, it decays into a pointer to the first element. So,

Code:

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void foo(char*);

int main(void){

    char myArray[]={'a','b','c'};
    foo(myArray);
    return (0);
}
void foo(char* array){
    char q;
    q=array[1];
    printf("%c", q); //this will print b
}

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Quote:

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Originally Posted by quzah

No, we just like to make up stuff here. :rolleyes:


Quzah.

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lolol

Thanks!

Quote:

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Originally Posted by quzah

No, we just like to make up stuff here. :rolleyes:
Quzah.

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Haha.....well sometimes we do, mostly when I post though. :wink::tongue: