How the values are accessed while passing an array string in a fucntion using a pointer?

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C++

#include<stdio.h>
#include<string.h>
void frequency_ptr(char *str);
int main()
{
    char strings[30];
    printf("Enter a string:");
    gets(strings);
    frequency_ptr(&strings[0]);
}
void frequency_ptr(char *strings)
{
    int i=0;
    static int cnt[26];
    while(*strings[i]!='\0')
    {
    if(*strings[i]>='a' && *strings[i]<='z')
    cnt[*strings[i]-'a']++;   
    i++;
    }
    for(i=0;i<26;i++)
        printf("\n%c occurs %d times:", 'a'+i, cnt[i]);
}

What I have tried:

The below program works fine for me. But why am i getting error in the above while loop.

C++

while(strings[i]!='\0')
 {
    if(strings[i]>='a' && strings[i]<='z')
    cnt[strings[i]-'a']++;
    i++;
 }

Thanks in advance.

Posted 31-Jul-18 11:16am

Member 13922884

Updated 31-Jul-18 23:15pm

KarstenK

v4

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Comments

Patrice T 31-Jul-18 17:35pm

'But why am i getting error in the above while loop.'
What error message ? and where ?

2 solutions

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CPallini · Accepted Answer · 2018-07-31T22:00:00

Solution 2

Don't ever use gets, use fgets instead. Try:

C

#include<stdio.h>
void frequency_ptr(const char *str);

#define SIZE 30
#define LETTERS 26

int main()
{
  char string[SIZE];
  printf("Enter a string:");
  fgets(string, sizeof(string), stdin);
  frequency_ptr(string);
  return 0;
}

void frequency_ptr(const char * str)
{
  unsigned int cnt[LETTERS]={0};
  while(*str != '\0')
  {
    if(*str>='a' && *str<='z')
      cnt[*str-'a']++;
    ++str;
  }
  size_t i;
  for(i=0; i<LETTERS;i++)
    printf("%c occurs %d times.\n", (char)('a'+i), cnt[i]);
}

Posted 31-Jul-18 22:00pm

CPallini

Comments

Member 13922884 1-Aug-18 5:09am

Thank you !!

CPallini 1-Aug-18 5:14am

You are welcome.

KarstenK 1-Aug-18 5:19am

By removing the i you changed the code more than needed. (Bad for comparisons) And I dont like changing input paramters. It is always confusing. ;-)

CPallini 1-Aug-18 5:59am

That is a common idiom with C-strings.
What you (and I) like doens't much matter. :-)

Jochen Arndt · Accepted Answer · 2018-07-31T11:48:00

The code at "What I have tried" is correct.

It is a common C/C++ beginners problem to understand the relation of pointers and arrays.

But it is covered by any C/C++ books and tutorials (search the web for something like "array vs pointer").

See for example the section "Pointers and Arrays" at C Class - Arrays, String Constants and Pointers[^].

In your case you have the function parameter char *strings. That means strings is a pointer to char values. Using the Dereference operator - Wikipedia[^], you would access the first character of the string:

C++

char first_char = *strings;

That can be also done with the array access [] operator:

C++

char first_char = strings[0];

So using

C++

char c = *strings[0];

is the same as

C++

char c = **strings;

and both result in a compilation error because the above would be only valid when strings is of type char ** (pointer to pointer to char).