Answer

In C, strings are implemented as arrays of characters, ending with a null byte ('\0'). Example of string declaration:

char str1[] = "hello";    // Array with 6 characters: {'h','e','l','l','o','\0'}
char *str2  = "hello";    // Pointer to a string literal

• Character array: Memory is allocated for each element, including the terminating null. The array can be modified (unless it is const).
• String pointer: Can point to a string literal (which cannot be modified) or to an allocated memory block. String literals are typically stored in the read-only segment.

To avoid errors:

• Properly manage memory (use malloc, strcpy, check buffer size).
• Do not use string literals for modifying strings.
• Ensure all strings are null-terminated \0.

Example of correct string handling:

char buffer[100];
strcpy(buffer, "test");    // OK, buffer is mutable and guaranteed to contain '\0'

Tricky Question

What is the result of executing the following code and what errors will it lead to?

char *str = "hello";
str[0] = 'H';
printf("%s\n", str);

Answer: The program will lead to undefined behavior, most likely a segmentation fault, because string literals are placed in read-only memory. Writing values at the address of a string literal is not allowed.

Examples of Real Errors Due to Lack of Knowledge

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Story The team confused the concepts of string array and pointer to literal. One function took char *output = "default"; and later executed strcpy(output, input);, which resulted in a crash on the first run, because the copy was occurring in read-only memory.

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Story When working with the network, the result was written to a buffer allocated with char *buf = NULL; strcpy(buf, data);. This led to writing to uninitialized memory and crashing the application.

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Story In a localization package, the team passed strings between components without ensuring that a \0 character was added. Once, the function printed garbage to the console and corrupted internal memory structure.