Consider these two statements: char v[6]; strcpy(v, "string");The best description of the situation after these two statements is that:Group of answer choicesThe call to strcpy() will corrupt memory and may result in a program run-time error at some later point in the program's executionA run-time array overflow will be detected, and the program will be halted and an error message printedThe call to strcpy() will be truncated, and only "tring" will be copied into vThe call to strcpy() is completely fine, and v will have the desired value of "string"The call to strcpy() will be truncated, and only "strin" will be copied into v
Question
Consider these two statements: char v[6]; strcpy(v, "string");The best description of the situation after these two statements is that:Group of answer choicesThe call to strcpy() will corrupt memory and may result in a program run-time error at some later point in the program's executionA run-time array overflow will be detected, and the program will be halted and an error message printedThe call to strcpy() will be truncated, and only "tring" will be copied into vThe call to strcpy() is completely fine, and v will have the desired value of "string"The call to strcpy() will be truncated, and only "strin" will be copied into v
Solution
The best description of the situation after these two statements is that: The call to strcpy() will corrupt memory and may result in a program run-time error at some later point in the program's execution.
Here's why:
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The statement
char v[6];declares an array of 6 characters. -
The statement
strcpy(v, "string");attempts to copy the string "string" into the array v. -
The string "string" actually contains 7 characters, not 6. This is because in C, strings are null-terminated, meaning they have a '\0' character at the end to signify the end of the string. So "string" is actually "string\0" in memory.
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Therefore, the strcpy() function will attempt to copy 7 characters into an array that only has space for 6. This will result in a buffer overflow, where the '\0' character will be written into memory that is not part of the array v.
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This can corrupt memory and may cause a run-time error later in the program's execution.
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