String to Integer atoi in DSA C

Mental Model

Convert each character digit in the string to a number and build the integer step by step.

Analogy: Like reading a number on a sign one digit at a time and writing it down to form the full number.

Input string: "  -42abc"
Positions:  [0] [1] [2] [3] [4] [5] [6]
Characters: ' '  '-'  '4'  '2'  'a'  'b'  'c'

Integer building:
result = 0
result = result * 10 + digit

Pointer ↑ at start of string

Dry Run Walkthrough

Input: string: " -42abc"

Goal: Convert the string to integer -42, ignoring trailing non-digit characters

Step 1: Skip leading spaces

Pointer ↑ at '-' (index 2), string: "  -42abc"

Why: Spaces do not affect the number, so we ignore them

Step 2: Check sign character '-' and set sign = -1

Pointer ↑ at '4' (index 3), sign = -1

Why: Minus sign means the number is negative

Step 3: Convert '4' to digit 4 and update result = 0 * 10 + 4 = 4

result = 4, pointer ↑ at '2' (index 4)

Why: First digit of the number

Step 4: Convert '2' to digit 2 and update result = 4 * 10 + 2 = 42

result = 42, pointer ↑ at 'a' (index 5)

Why: Second digit of the number

Step 5: Encounter non-digit 'a', stop parsing

result = 42, pointer at 'a' (index 5)

Why: Non-digit means number ends here

Step 6: Apply sign to result: result = -1 * 42 = -42

final result = -42

Why: Sign affects the final integer value

Result:

final integer = -42

Annotated Code

DSA C

#include <stdio.h>
#include <limits.h>

int myAtoi(const char *str) {
    int i = 0;
    int sign = 1;
    long result = 0;

    // Skip leading spaces
    while (str[i] == ' ') {
        i++;
    }

    // Check sign
    if (str[i] == '-' || str[i] == '+') {
        sign = (str[i] == '-') ? -1 : 1;
        i++;
    }

    // Convert digits to integer
    while (str[i] >= '0' && str[i] <= '9') {
        result = result * 10 + (str[i] - '0');

        // Check overflow
        if (sign == 1 && result > INT_MAX) {
            return INT_MAX;
        }
        if (sign == -1 && -result < INT_MIN) {
            return INT_MIN;
        }
        i++;
    }

    return (int)(sign * result);
}

int main() {
    const char *input = "  -42abc";
    int output = myAtoi(input);
    printf("%d\n", output);
    return 0;
}

while (str[i] == ' ') { i++; }

skip all spaces before number

if (str[i] == '-' || str[i] == '+') { sign = (str[i] == '-') ? -1 : 1; i++; }

detect and record sign, move pointer forward

while (str[i] >= '0' && str[i] <= '9') {

process each digit character

result = result * 10 + (str[i] - '0');

build integer by shifting digits left and adding new digit

if (sign == 1 && result > INT_MAX) { return INT_MAX; }

handle positive overflow by clamping

if (sign == -1 && -result < INT_MIN) { return INT_MIN; }

handle negative overflow by clamping

return (int)(sign * result);

apply sign and return final integer

Output

-42

Complexity Analysis

Time: O(n) because we scan the string once until non-digit or end

Space: O(1) because we use fixed variables, no extra data structures

vs Alternative: Compared to using built-in functions, this manual parsing avoids overhead and allows custom overflow handling

Edge Cases

empty string

returns 0 because no digits found

DSA C

while (str[i] >= '0' && str[i] <= '9') {

string with only spaces

returns 0 after skipping spaces and finding no digits

DSA C

while (str[i] == ' ') { i++; }

string with number exceeding int max

returns INT_MAX to avoid overflow

DSA C

if (sign == 1 && result > INT_MAX) { return INT_MAX; }

string with number below int min

returns INT_MIN to avoid overflow

DSA C

if (sign == -1 && -result < INT_MIN) { return INT_MIN; }

string with no digits after sign

returns 0 because no digits parsed

DSA C

while (str[i] >= '0' && str[i] <= '9') {

When to Use This Pattern

When you see a problem asking to convert a string to an integer with possible spaces and signs, use the atoi pattern to parse character by character and handle edge cases.

Common Mistakes

Mistake: Not skipping leading spaces before parsing digits
Fix: Add a loop to skip spaces before checking sign or digits

Mistake: Not handling sign character correctly
Fix: Check for '+' or '-' and set sign variable accordingly before parsing digits

Mistake: Not stopping parsing at first non-digit character
Fix: Stop the digit parsing loop when a non-digit is encountered

Mistake: Ignoring integer overflow and underflow
Fix: Check if result exceeds INT_MAX or INT_MIN and clamp accordingly

Summary

Converts a string to an integer by reading digits and applying sign.

Use when you need to parse numbers from strings with spaces and signs.

The key is to process characters one by one, handle sign, and stop at first non-digit.