In C, what will this program print?\n\n#include<stdio.h>\n#include<stdlib.h>\n\nint main()\n{\n char *i = "55.555";\n int result1 = 10;\n float result2 = 11.111f;\n result1 = result1 + atoi(i);\n result2 = result2 + atof(i);\n printf("%d, %f", result1, result2);\n return 0;\n}\n\nAssume default float formatting with six digits after the decimal.

Introduction / Context:
atoi converts a string to int, stopping at the first non-digit. atof converts a string to double (then may be assigned to float). Understanding how each treats a string like "55.555" explains the different results for integer and floating additions.

Given Data / Assumptions:

• atoi("55.555") → 55 (parses leading digits only).
• atof("55.555") → 55.555.
• result1 starts at 10; result2 starts at 11.111.
• printf("%d, %f") prints the int and a floating value with six decimals.

Concept / Approach:
Compute each sum separately using the conversions provided by atoi and atof.

Step-by-Step Solution:
result1 = 10 + atoi("55.555") = 10 + 55 = 65.result2 = 11.111 + atof("55.555") = 11.111 + 55.555 = 66.666.printf prints the float as 66.666000 (six fractional digits by default).

Verification / Alternative check:
Replacing atoi with strtol and atof with strtod yields the same numeric results for this input.

Why Other Options Are Wrong:
55, 55.555: ignores the initial 10 and 11.111.66, 66.666600: integer sum is off by 1.55, 55: both values disregard conversion rules and initial values.

Common Pitfalls:
Expecting atoi to round or parse decimals; confusing float print precision.

Final Answer:
65, 66.666000