pointer arithmetic in C for getting a string

Question

I want to get the elements of an array of characters, but no success at all, the problem is that I only get the first and last element and nothing more, my code is:

void getcharacters(char *cad)
{
 int l;
 int *i;
 l=strlen(cad);
 for (i=&cad[0];i<&cad[l];i++){
     printf("%c\n",*cad);
 }
}

any help? Thanks

Also, why print *cad? It never changes from the first character. — goji
– goji, Commented Dec 13, 2012 at 3:41

Karthik T · Accepted Answer · 2012-12-13 04:46:36Z

4

The size of an int can be as big as 4 x the size of a char, so when you do i++ you are actually skipping 3 chars.

Also, you print out *cad instead of *i.

To fix change i to char* and print *i instead of *cad

edited Dec 13, 2012 at 4:46

answered Dec 13, 2012 at 3:42

Karthik T

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2 Comments

Alvin Wong Over a year ago

"int can be as big as 4 chars" lol, I would say char is not guaranteed to be the same size of int (and in most cases they will never be)

Karthik T Over a year ago

@AlvinWong typically sizeof(int) == 4 while sizeof(char) == 1 i believe

perreal · Accepted Answer · 2012-12-13 03:43:07Z

2

You are using the same character (*cad or cad[0]) for all printf's. What you need is to use the index to get the next char in each iteration. Also i needs to be a pointer to char:

void getcharacters(char *cad)
{
 int l;
 char *i;
 l=strlen(cad);
 for (i=&cad[0];i<&cad[l];i++){
     printf("%c\n", *i );
 }
}

answered Dec 13, 2012 at 3:43

perreal

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Comments

Templar · Accepted Answer · 2012-12-13 03:43:36Z

2

Why don't you iterate from first character to the last one and accessing them as array index

int i;
int l=strlen(cad);
for (i=0;i<l;i++)
{
  printf("%c\n",cad[i]);
}

answered Dec 13, 2012 at 3:43

Templar

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Comments

Templar · Accepted Answer · 2012-12-13 03:54:08Z

1

Other answers have told you why it plain doesn't work, I'm wonder why you're not just iterating until the null terminator?

void getcharacters(char *cad)
{
 char *i;
 for (i = cad; *i; i++) {
     printf("%c\n",*i);
 }
}

edited Dec 13, 2012 at 3:54

Templar

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answered Dec 13, 2012 at 3:46

goji

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1 Comment

Templar Over a year ago

i should be a *char instead of *int

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