Convert String to Integer in C++

How to convert a string to integer in C?

There is strtol which is better IMO. Also I have taken a liking in strtonum, so use it if you have it (but remember it's not portable):

long long
     strtonum(const char *nptr, long long minval, long long maxval,
     const char **errstr);

You might also be interested in strtoumax and strtoimax which are standard functions in C99. For example you could say:

uintmax_t num = strtoumax(s, NULL, 10);
if (num == UINTMAX_MAX && errno == ERANGE)
    /* Could not convert. */

Anyway, stay away from atoi:

The call atoi(str) shall be equivalent to:

(int) strtol(str, (char **)NULL, 10)

except that the handling of errors may differ. If the value cannot be
represented, the behavior is undefined.

How to convert an int to string in C?

EDIT: As pointed out in the comment, itoa() is not a standard, so better use sprintf() approach suggested in the rivaling answer!

---

You can use itoa() function to convert your integer value to a string.

Here is an example:

int num = 321;
char snum[5];

// convert 123 to string [buf]
itoa(num, snum, 10);

// print our string
printf("%s\n", snum);

---

If you want to output your structure into a file there is no need to convert any value beforehand. You can just use the printf format specification to indicate how to output your values and use any of the operators from printf family to output your data.

Converting gets() string into an integer in C

Instead of using your own method to do this, there are a number of utilities in the standard library. Take a look at strol and sscanf. It's also wise to use fgets instead of gets as pointed out in the comments above.

Example

#include <stdio.h>
#include <stdlib.h>
int main()
{
    char s[1000];
    int n = 0;
    printf("Input the number you wish to have converted\n");//asks the user to enter the number they want converted
    fgets(s, sizeof(s), stdin);//reads the input

    n = (int)strol(s, NULL, 10);
    printf("Number from strol: %d\n", n);

    sscanf(s, "%d", &n);
    printf("Number from sscanf: %d\n", n);
}

You can even bypass fgets and use scanf if you don't want to keep the string:

#include <stdio.h>
int main()
{
    int n;
    scanf("%d", &n);
    printf("Number from scanf: %d\n", n);
}

Convert string into number

Yes. You can use strtol function.

long int strtol(const char * restrict nptr, char ** restrict endptr, int base);  

The strtol convert the initial portion of the string pointed to by nptr to long int representation.

Better to not use atoi. It tells noting when unable to convert a string to integer unlike strtol which specify by using endptr that whether the conversion is successful or not. If no conversion could be performed, zero is returned.

Suggested reading: correct usage of strtol

Example:

char *end;
char *str = "test";
long int result = strtol(str, &end, 10);  

if (end == str || *temp != '\0')
    printf("Could not convert '%s' to long and leftover string is: '%s'\n", str, end);  
else 
    printf("Converted string is: %ld\n", result);

How to convert character string to int in c

There is no need to use the function atoi. Instead of this code snippet

len = strlen(str);
for(i=0; i<len; i++)
{
  sum= sum + atoi(&str[i]);
}

you could write

for ( const char *p = str; *p; ++p )
{
    if ( '0' < *p && *p <= '9' ) sum += *p - '0';
}

If to use your approach with the function atoi then for example the string 123 is parsed as 123 + 23 + 3 = 149.

how can I split and convert string to integer?

You can use sscanf:

char* input = "1999-12-05";
int year;
int month;
int day;
sscanf(input, "%d-%d-%d", &year, &month, &day);
printf("year: %d, month: %d, day: %d\n", year, month, day);

demo

C program to convert string to integer and solve them algebraically

Your code has syntax errors.

• You forgot the " around %s
• you did not declare S

Here is a simpler solution:

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

int main(void) {
    char buf[200];

    if (fgets(buf, sizeof buf, stdin)) {
        long sum = 0;

        /* remove all white space */
        for (int i = j = 0, len = strlen(buf); i <= len; i++) {
            if (!isspace((unsigned char)buf[i]))
                buf[j++] = buf[i];
        }
        /* parse the expression */
        for (char *p = buf, *q;;) {
            long a = strtol(p, &q, 10);
            if (p == q) {
                break;
            } else {
                sum += a;
                p = q;
            }
        }
        printf("sum = %ld\n", sum);
    }
    return 0;
}

convert string to integer and getback same string from integer value in c

There is absolutely no way to convert a string to an int and then back again the way you want.

An int is typically 4 bytes, and you're using a char[50] which is 50 bytes. This means that an int can have 2³² different values, while a char[50] can have 2⁴⁰⁰ different values. So it is simply impossible to map them one to one.

Let's take an example outside the realm of code and computers and just focus on numbers. Can you imagine a method to convert a two-digit number to a one-digit number and back? If this was possible, then we would not need two digits in the first place. If such a method existed, you would be able to store an infinite amount of data in a single bit.

You can convert a char[4] to an int and back. It's actually really easy. This code will print abcd.

char str[4] = { 'a', 'b', 'c', 'd'};
// Convert to int
int result = *(int*) str;

char newstr[4];

// Convert back
char * ptr = (char*) &result;
for(int i=0; i<4; i++)
    newstr[i] = ptr[i];

printf("%.4s\n", newstr);

Note that I completely ignored termination of the string, and just instructed printf to stop after four characters.

Oh, and never use the gets function. It's dangerous.

---

Related Topics