0
221
ViewsPointers and Arrays in C, Need for more Understanding
I was doing some pointers and arrays practice in C and I noticed all my four methods returned the same answer.My question is are there disadvantages of using any one of my below methods? I am stunned at how all these four give me the same output. I just noticed you can use a pointer as if it was an array and you can also use an array as if it was a pointer?
char *name = "Madonah";
int i= 0;
for (i=0;i<7; i++){
printf("%c", *(name+i));
}
char name1 [7] = "Madonah";
printf("\n");
int j= 0;
for (j=0;j<7; j++){
printf("%c", name1[j]);
}
char *name2 = "Madonah";
printf("\n");
int k= 0;
for (k=0;k<7; k++){
printf("%c", name2[k]);
}
char name3 [7] = "Madonah";
printf("\n");
int m= 0;
for (m=0;m<7; m++){
printf("%c", *(name+m));
}
Results:
Madonah
Madonah
Madonah
Madonah
3 answers
Answer question0
It is true that pointers and arrays are equivalent in some context, "equivalent" means neither that they are identical nor even interchangeable. Arrays are not pointers.
It is pointer arithmetic and array indexing that are equivalent, pointers and arrays are different.
which one is preferable and the advantages/Disadvantages?
It depends how you want to use them. If you do not wanna modify string then you can use
char *name = "Madonah";
It is basically equivalent to
char const *name = "Madonah";
*(name + i) and name[i] both are same. I prefer name[i] over *(name + i) as it is neat and used most frequently by C and C++ programmers.
If you like to modify the string then you can go with
char name1[] = "Madonah";
0
In C, a[b], b[a], *(a+b) are equivalent and there's no difference between these 3. So you only have 2 cases:
char *name = "Madonah"; /* case 1 */
and
char name3 [7] = "Madonah"; /* case 2 */
The former is a pointer which points to a string literal. The latter is an array of 7 characters.
which one is preferred depends on your usage of name3.
If you don't intend to modify then the string then you can use (1) and I would also make it const char* to make it clear and ensure the string literal is not modified accidentally. Modifying string literal is undefined behaviour in C
If you do need to modify it then (2) should be used as it's an array of characters that you can modify. One thing to note is that in case (2), you have explicitly specified the size of the array as 7. That means the character array name3 doesn't have a null terminator (\0) at the end. So it can't be used as a string. I would rather not specify the size of the array and let the compiler calculate it:
char name3 [] = "Madonah"; /* case 2 */
/* an array of 8 characters */
0
Just in addition to what others said, I will add an image for better illustration. If you have
char a[] = "hello";
char *p = "world";
What happens in first case enough memory is allocated for a (6 characters) on the stack usually, and the string "hello" is copied to memory which starts at a. Hence, you can modify this memory region.
In the second case "world" is allocated somewhere else(usually in read only region), and a pointer to that memory is returned which is simply stored in p. You can't modify the string literal in this case via p.
Here is how it looks:
But for your question stick to notation which is easier, I prefer [].
More info on relationship between arrays and pointers is here.