Question

Given the following C program, #include int main(int argc, char *argv[]) { char str1[8] = "START"; char str2[8]; gets(str2); printf("str1 = %s, str2 = %s\\", str1, str2); return 0; } if the user enters Hello, world! for str2, then the value of str1 would be _____ and the value of str2 would be _____.

Answer 1

The value of str1 would be 'STARHell' and the value of str2 would be 'Hello, world!'.

Step-by-step explanation:

The given C program declares two character arrays, str1 and str2. str1 is initialized with the value 'START' and str2 is left uninitialized. The gets() function is used to read input from the user and store it in str2. However, gets() is a dangerous function and can lead to buffer overflow vulnerabilities.

If the user enters 'Hello, world!' for str2, the null character at the end of 'Hello' will overwrite the 'T' in str1, resulting in the values of str1 and str2 becoming 'STARHell' and 'Hello, world!', respectively.

It is important to note that using gets() is highly discouraged due to its security risks. Instead, the fgets() function should be used with appropriate buffer size.

Answer 2

str1 would be overwritten with some part of "Hello, world!" starting from 'w', potentially followed by some original characters from "START".

str2 would contain the truncated string "Hello, w".

The given C program uses the gets function to read a string from the user and store it in the array str2. The gets function is unsafe because it does not check the length of the input and can cause a buffer overflow if the user enters more characters than the array can hold.

In the program, both str1 and str2 are character arrays (buffers) of size 8. str1 is initialized with the string "START", which with the null terminator '\0' takes up 6 bytes out of the 8 available.

If the user enters "Hello, world!" for str2, this input exceeds the size of str2 which can only safely contain 7 characters plus the null terminator. Since gets does not perform bounds checking, the excess characters can overflow into adjacent memory. In this case, str2 directly follows str1 in memory, so the overflow could overwrite the contents of str1.

The input "Hello, world!" is 13 characters long plus the null terminator, which is a total of 14 bytes. This will completely fill str2's 8 bytes and then overwrite the next 6 bytes, which happen to be the storage for str1. Since the memory layout in C is not guaranteed and can vary by compiler and architecture, the exact result after the overflow could be unpredictable. However, typically in such cases, str1 would be overwritten starting from its first byte with the overflowed bytes from str2.

Therefore, after the overflow, str1 would contain the beginning of the string "Hello, world!" starting from the 9th character ('w') followed by whatever fits into the remaining space of str1. Since the input is "Hello, world!" and str1 can be overwritten from the 9th character, str1 will likely start with "world!" and a couple of characters from the end of "START". However, it's also possible that due to memory alignment or other factors, the overflow could have a different effect.

As for str2, it would contain "Hello, world!" up to the point of buffer overflow, truncated to fit the 8-byte space, which would be "Hello, w".

str1 would be overwritten with some part of "Hello, world!" starting from 'w', potentially followed by some original characters from "START".

str2 would contain the truncated string "Hello, w".