Parsing numeric arguments - Interactive Code Practice

Practice - 5 Tasks

1fill in blank

easy

Complete the code to convert the string argument to an integer using atoi.

C

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

int main(int argc, char *argv[]) {
    if (argc < 2) {
        printf("Please provide a number.\n");
        return 1;
    }
    int num = [1](argv[1]);
    printf("Number is %d\n", num);
    return 0;
}

Drag options to blanks, or click blank then click option'

Aatoi

Bstrlen

Cprintf

Dscanf

Attempts:

3 left

💡 Hint

Common Mistakes

Using printf instead of atoi.

Using strlen which returns string length, not number.

Using scanf incorrectly here.

The function atoi converts a string to an integer. Here, argv[1] is the string argument to convert.

2fill in blank

medium

Complete the code to parse a long integer from the string argument using strtol.

C

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

int main(int argc, char *argv[]) {
    if (argc < 2) {
        printf("Please provide a number.\n");
        return 1;
    }
    char *endptr;
    long val = [1](argv[1], &endptr, 10);
    if (*endptr != '\0') {
        printf("Invalid number: %s\n", argv[1]);
        return 1;
    }
    printf("Parsed long: %ld\n", val);
    return 0;
}

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Astrcpy

Bstrlen

Catoi

Dstrtol

Attempts:

3 left

💡 Hint

Common Mistakes

Using atoi which does not detect invalid characters.

Using strcpy which copies strings but does not convert.

Using strlen which returns length, not number.

strtol converts a string to a long integer and allows error checking via endptr.

3fill in blank

hard

Fix the error in the code to correctly parse a double from the string argument using strtod.

C

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

int main(int argc, char *argv[]) {
    if (argc < 2) {
        printf("Please provide a number.\n");
        return 1;
    }
    char *endptr;
    double val = [1](argv[1], &endptr);
    if (*endptr != '\0') {
        printf("Invalid number: %s\n", argv[1]);
        return 1;
    }
    printf("Parsed double: %f\n", val);
    return 0;
}

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Astrtol

Bstrtod

Catoi

Datof

Attempts:

3 left

💡 Hint

Common Mistakes

Using atof which does not allow error checking.

Using atoi which converts to int, not double.

Using strtol which converts to long, not double.

strtod converts a string to a double and allows error checking via endptr.

4fill in blank

hard

Fill both blanks to create a dictionary-like structure that maps argument index to its parsed integer value if the argument is numeric.

C

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

int is_number(const char *s) {
    while (*s) {
        if (![2](*s)) return 0;
        s++;
    }
    return 1;
}

int main(int argc, char *argv[]) {
    for (int i = 1; i < argc; i++) {
        if (is_number(argv[i])) {
            int val = [1](argv[i]);
            printf("Arg %d is number %d\n", i, val);
        } else {
            printf("Arg %d is not a number\n", i);
        }
    }
    return 0;
}

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Aatoi

Bstrtol

Cisdigit

Dprintf

Attempts:

3 left

💡 Hint

Common Mistakes

Using strtol without proper error checking.

Using printf instead of conversion function.

Using isdigit on whole string instead of characters.

atoi converts string to int; isdigit checks if a character is a digit.

5fill in blank

hard

Fill all three blanks to parse command line arguments as integers, check for errors, and print the sum.

C

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

int main(int argc, char *argv[]) {
    int sum = 0;
    for (int i = 1; i < argc; i++) {
        char *endptr;
        int val = (int)[1](argv[i], &[2], 10);
        if (*[3] != '\0') {
            printf("Invalid number: %s\n", argv[i]);
            return 1;
        }
        sum += val;
    }
    printf("Sum is %d\n", sum);
    return 0;
}

Drag options to blanks, or click blank then click option'

Astrtol

Bendptr

Datoi

Attempts:

3 left

💡 Hint

Common Mistakes

Using atoi which does not allow error checking.

Not checking endptr to detect invalid input.

Confusing variable names for endptr.

strtol parses string to long with error checking via endptr pointer.

Practice

1. What does the atoi function do in C when parsing numeric arguments?

easy

A. Converts a string to an integer without error checking

B. Converts a string to a floating-point number

C. Checks if a string contains only digits

D. Parses a string and returns a pointer to the next character

Solution

1. Step 1: Understand atoi function purpose

   atoi converts a string to an integer but does not check for errors or invalid input.

2. Step 2: Compare with other parsing functions

   Unlike strtol, atoi does not provide error detection or pointer to next character.

3. Final Answer:

   Converts a string to an integer without error checking -> Option A

4. Quick Check:

   atoi = simple integer conversion [OK]

Hint: Remember: atoi converts but doesn't check errors [OK]

Common Mistakes:

• Thinking atoi checks for invalid input
• Confusing atoi with strtol
• Assuming atoi returns pointer to next char

2. Which of the following is the correct syntax to parse a string "1234" to an integer using strtol in C?

easy

A. int num = strtol("1234", NULL, 10);

B. long num = strtol("1234", NULL, 10);

C. long num = atoi("1234", NULL, 10);

D. int num = strtol("1234", 10, NULL);

Solution

1. Step 1: Check strtol return type

   strtol returns a long, so the variable must be long.

2. Step 2: Verify argument order

   Arguments are: string, pointer to char* for end, and base (10 for decimal). So strtol("1234", NULL, 10) is correct.

3. Final Answer:

   long num = strtol("1234", NULL, 10); -> Option B

4. Quick Check:

   strtol returns long, args: string, endptr, base [OK]

Hint: Remember strtol returns long, not int [OK]

Common Mistakes:

• Using int instead of long for strtol result
• Mixing argument order in strtol
• Using atoi syntax with strtol

3. What will be the output of this C code snippet?

#include 
#include 

int main() {
    char *str = "42abc";
    char *endptr;
    long val = strtol(str, &endptr, 10);
    printf("%ld %s\n", val, endptr);
    return 0;
}

medium

A. Error

B. 0 42abc

C. 42abc (null)

D. 42 abc

Solution

1. Step 1: Understand strtol parsing behavior

   strtol converts initial numeric part "42" to long and sets endptr to point after parsed number, which is "abc".

2. Step 2: Analyze printf output

   It prints the number 42 and then the string starting at endptr, which is "abc".

3. Final Answer:

   42 abc -> Option D

4. Quick Check:

   Parsed number + leftover string = 42 abc [OK]

Hint: Check where endptr points after strtol [OK]

Common Mistakes:

• Assuming strtol parses entire string
• Ignoring endptr usage
• Expecting error on non-digit suffix

4. Identify the error in this code snippet that parses a numeric argument:

#include 

int main(int argc, char *argv[]) {
    int num = atoi(argv[1]);
    if (num == 0) {
        printf("Invalid number\n");
    }
    return 0;
}

medium

A. Missing check if argc > 1 before accessing argv[1]

B. Using int instead of long for atoi result

C. atoi returns 0 for invalid input and valid zero, causing confusion

D. No error, code is correct

Solution

1. Step 1: Check argument count before access

   The code uses argv[1] without verifying argc > 1, risking out-of-bounds access.

2. Step 2: Understand atoi zero ambiguity

   While atoi returns 0 for invalid input and valid zero, the bigger issue is missing argument check causing runtime error.

3. Final Answer:

   Missing check if argc > 1 before accessing argv[1] -> Option A

4. Quick Check:

   Always check argc before argv access [OK]

Hint: Always check argc before argv usage [OK]

Common Mistakes:

• Ignoring argc check leads to crashes
• Confusing atoi zero return meaning
• Assuming atoi handles errors

5. You want to safely parse a command-line argument as a positive integer in C. Which approach correctly handles invalid input and zero?

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

int main(int argc, char *argv[]) {
    if (argc <= 1) {
        printf("No input provided\n");
        return 1;
    }
    char *endptr;
    long val = strtol(argv[1], &endptr, 10);
    if (*endptr != '\0' || val <= 0) {
        printf("Invalid positive integer\n");
        return 1;
    }
    printf("Parsed value: %ld\n", val);
    return 0;
}

hard

A. Fails to check if argv[1] is NULL before parsing

B. Does not handle negative numbers correctly

C. Correctly parses positive integers and rejects invalid or zero input

D. Should use atoi instead of strtol for safety

Solution

1. Step 1: Check argument count

   The code checks if argc <= 1 to ensure input is provided.

2. Step 2: Use strtol with endptr to detect invalid characters and check if value is positive

   It verifies *endptr == '\0' to confirm full string parsed and val > 0 to ensure positive integer.

3. Final Answer:

   Correctly parses positive integers and rejects invalid or zero input -> Option C

4. Quick Check:

   strtol + endptr + value check = safe parsing [OK]

Hint: Use strtol with endptr and check value > 0 [OK]

Common Mistakes:

• Using atoi which can't detect invalid input
• Not checking endptr for leftover chars
• Ignoring zero or negative values