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      Question

      Consider the following C code: #include

      #include int main() {     char str1[] = "Hello";     char str2[] = {'W', 'o', 'r', 'l', 'd', '\0', 'X'};     char str3[] = {'C', 'o', 'd', 'e'}; // No null terminator     printf("%zu %zu\n", strlen(str1), strlen(str2));     // What happens if strlen(str3) is called?     // printf("%zu\n", strlen(str3));     return 0; } What will be the output of the printf statement, and what would be the behavior if strlen(str3) were uncommented and executed?
      A 5 5; strlen(str3) would print 4. Correct Answer Incorrect Answer
      B 5 5; strlen(str3) would cause a segmentation fault or print a garbage value. Correct Answer Incorrect Answer
      C 5 6; strlen(str3) would print 4. Correct Answer Incorrect Answer
      D 5 6; strlen(str3) would cause a segmentation fault or print a garbage value. Correct Answer Incorrect Answer
      E 5 7; strlen(str3) would print 4. Correct Answer Incorrect Answer

      Solution

      Correct Answer: B (strlen(str1) is 5. strlen(str2) counts up to the first \0, so 5. strlen(str3) would read past the allocated memory until it finds a \0, leading to undefined behavior (segmentation fault or garbage value).)

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