Parsing numeric arguments - Practice Problems & Coding Challenges

Challenge - 5 Problems

❓ Predict Output

intermediate

Output of parsing integer with strtol

What is the output of this C program when parsing the string "123abc" using strtol?

C

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

int main() {
    const char *str = "123abc";
    char *endptr;
    long val = strtol(str, &endptr, 10);
    printf("Value: %ld\n", val);
    printf("Remaining string: %s\n", endptr);
    return 0;
}

A

Value: 123
Remaining string: abc

B

Value: 0
Remaining string: 123abc

C

Value: 0
Remaining string: 

D

Value: 123
Remaining string: 

💡 Hint

Remember that strtol stops parsing when it encounters a non-digit character.

The function strtol parses the initial part of the string as a number until it finds a character that is not valid for the number base. Here, it parses "123" and stops at "a", so val is 123 and endptr points to "abc".

❓ Predict Output

intermediate

Output when parsing invalid number with strtol

What is the output of this C program when parsing the string "abc123" using strtol?

C

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

int main() {
    const char *str = "abc123";
    char *endptr;
    long val = strtol(str, &endptr, 10);
    if (endptr == str) {
        printf("No digits were found\n");
    } else {
        printf("Value: %ld\n", val);
    }
    return 0;
}

A

No digits were found

B

Value: 0

C

Value: -1

D

Value: 123

💡 Hint

Check what happens if the string does not start with digits.

strtol returns 0 and sets endptr to the original string if no digits are found at the start. The program checks if endptr equals str to detect this case and prints "No digits were found".

❓ Predict Output

advanced

Output of parsing hexadecimal number with strtol

What is the output of this C program when parsing the string "0x1A3f" using strtol with base 0?

C

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

int main() {
    const char *str = "0x1A3f";
    char *endptr;
    long val = strtol(str, &endptr, 0);
    printf("Value: %ld\n", val);
    printf("Remaining string: %s\n", endptr);
    return 0;
}

A

Value: 0
Remaining string: 0x1A3f

B

Value: 6719
Remaining string: 

C

Value: 1
Remaining string: A3f

D

Value: 6719
Remaining string: 0x1A3f

💡 Hint

Base 0 lets strtol detect the base from the string prefix.

When base is 0, strtol detects the base automatically. The prefix "0x" means hexadecimal, so it parses "1A3f" as hex which is 6719 decimal. The entire string is parsed, so endptr points to the empty string.

❓ Predict Output

advanced

Output when parsing float with strtol

What is the output of this C program when parsing the string "3.14" using strtol?

C

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

int main() {
    const char *str = "3.14";
    char *endptr;
    long val = strtol(str, &endptr, 10);
    printf("Value: %ld\n", val);
    printf("Remaining string: %s\n", endptr);
    return 0;
}

A

Value: 3
Remaining string: 14

B

Value: 0
Remaining string: 3.14

C

Value: 3
Remaining string: .14

D

Value: 314
Remaining string: 

💡 Hint

strtol stops parsing at the first character that is not valid for an integer.

strtol parses the initial integer part "3" and stops at the dot "." which is not valid for integers. So val is 3 and endptr points to ".14".

🧠 Conceptual

expert

Behavior of strtol on overflow

What happens when strtol parses a number larger than LONG_MAX in C?

A It causes a runtime crash due to overflow.

B It returns 0 and sets errno to EINVAL.

C It returns the number modulo LONG_MAX without error.

D It returns LONG_MAX and sets errno to ERANGE.

💡 Hint

Check the C standard behavior for overflow in strtol.

When the parsed number is out of range for long, strtol returns LONG_MAX or LONG_MIN depending on sign and sets errno to ERANGE to indicate overflow.