CODEWITHSUNDEEP
carraysstring
konde020202
konde020202

Reputation: 27

How do I convert string to integer in this case?

I have this problem as a part of my c-programming project. I read users input to char type array (char*str) and I need to convert some parts of the string input to integer. The input might be "A smeagol 21 fire 22"

Here is some testing. I try to get x=40. This code's gives x=-4324242. Why this code don't work?

#include <stdio.h>

int main(){
    char *uga[1];
    uga[0] = "10";
    printf("%s\n", uga[0]);
    int x = 50 - (int)uga[0];
    printf("%d",x);     
}

Thank you beforehand.

Upvotes: 1

Views: 748

Answers (5)

Prateek Tewary
Prateek Tewary

Reputation: 33

You should use the function atoi() under stdlib.h.
For instance,

    int val; 
    char strn1[] = "12546"; 
  
    val = atoi(strn1); 
    printf("String value = %s\n", strn1); 
    printf("Integer value = %d\n", val); 
  
    char strn2[] = "12Ge 1 89"; 
    val = atoi(strn2); 
    printf("String value = %s\n", strn2); 
    printf("Integer value = %d\n", val);

Output:

String value = 12546
Integer value = 12546
String value = 12Ge 1 89
Integer value = 12

Hope it helps! gives the

Upvotes: 0

Hitokiri
Hitokiri

Reputation: 3699

You can use strtol, sscanf or atoi to convert a string to int. For example:

#include <stdlib.h>
#include <stdio.h>  
#include <string.h>

int main() {
    char *str[3] = {"10", "20 test of strtol\n", "30"};
    int a = atoi(str[0]);
    printf("a = %d\n", a);
    char *ptr;
    long int b = strtol(str[1], &ptr, 10); // you can alse use strtoll for long long int
    printf("b = %ld, string: %s", b, ptr);
    int c;
    sscanf(str[2], "%d", &c);
    printf("c = %d\n", c);
    return 0;
}

Output:

a = 10
b = 20, string:  test of strtol
c = 30

In your code, uga[0] is a pointer that points to a character. So (int) uga[0] just cast the address of the char pointer.

Upvotes: 1

Eraklon
Eraklon

Reputation: 4288

You can use sscanf also like this for example

#include <stdio.h>

int main(){
    char str[] = "A smeagol 21 fire 22";
    int a, b;

    if (2 != sscanf(str, "%*[^0-9] %d %*[^0-9] %d", &a, &b))
      return printf("Error\n"), 1;

    printf("%d %d",a, b);     
}

Output

21 22

%*[^0-9] will match any non digit character and it will be discarded (because of the *). sscanf return the number of variables which it was able to fill. Which should be 2 here, otherwise something went wrong.

Upvotes: 0

Lahcen YAMOUN
Lahcen YAMOUN

Reputation: 691

There's two problems in your code.

  1. uga isn't a string as you think, but an array of strings containing only one string that has no memory allocated. To fix that, you should declare uga as it follows char uga[max_size_of_your_string] = "Initialization if you want to";
  2. Thus (int)uga[0]; won't work as you expect, because you're trying to convert an address to an integer.

I advise you to declare properly your string uga, and to take a look at the different ways of converting a numerical string to integer here : How to convert a string to integer in C?

Upvotes: 0

Govind Parmar
Govind Parmar

Reputation: 21542

Type casting a char pointer or array to int or another numeric type will not give you the numeric value of the text in the string.

You should use the functions strtol (signed), strtoul (unsigned) to accomplish this:

char *str = "123ABC";
char *pStop;     

// (Arguments: string to parse, pointer to character that stops the parse, base/radix)
int value = strtol(str, &pStop, 10);

After this, value will be 123, and pStop will point to the 'A' in str. The second argument can be NULL if you don't care about it, but it's useful for processing a string with lots of numbers in it between non-numeric text.

Upvotes: 0

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