Pointer to array of string

Question

I don't understand this part of code:

char *arrStr[3];
arrStr[0] = "hola";    //works
*(arrStr+1) = "guys";  //works
arrStr++;              // doesn't work why?

char **arrStr2 = arrStr; //works
arrStr2++; //works

I don't understand why arrStr++ doesn't work while arrStr2 works.

This is strange because I'm doing the same thing, but in one case it works in the other not, why is that?

haccks · Accepted Answer · 2013-11-29 02:27:46Z

3

*arrStr[3] is an array of 3 pointers to char while **arrStr2 a pointer to a pointer to char. In

arrStr[0] = "hola";

arrStr is an array and array names are nonmodifiable l-values and hence you can't modify it. On the other hand modification can be done on arrStr as it is a pointer to pointer but not an array. Do remember that arrays are not pointers.

edited Nov 29, 2013 at 2:27

answered Nov 29, 2013 at 2:21

haccks

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Yu Hao · Accepted Answer · 2013-11-29 02:21:44Z

2

arrStr is an array of pointers, you can't modify an array name. arrStr++; won't compile.

arrStr2 is a pointer to a pointer, you can modify a pointer. And you should consider a better name, as it's not an array.

answered Nov 29, 2013 at 2:21

Yu Hao

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3 Comments

Yu Hao Over a year ago

@GiovanniFar An array name is automatically converted to a pointer to the first element in some situations(like passed as function argument), but no, arrays are not pointers.

Giovanni Far Over a year ago

ok but both have a reference ... so i cannot see the difference

Giovanni Far Over a year ago

thanks so much. now is clear. with an array i'm moving direclty inside the array, with the point i change just target and after i fetch it. but all the process is inside to the reference of the pointer. with the array no... i have a reference but i go direclty there and i move position there. with the pointer you remain inside it and increment there the reference and in the end you go to fetch the value inside at this reference

haccks · Accepted Answer · 2013-11-29 02:31:18Z

0

char *arrStr[3];

if you use the array name as a pointer,you must know that this pointer is a const pointer, so you can't do the ++ or -- operator, char *str = "ciao"; just declare a normal pointer, you can do every to it.

edited Nov 29, 2013 at 2:31

haccks

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answered Nov 29, 2013 at 2:20

segment_fault

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