Intuition The ONE core ideaA C "string" is just a row of bytes in memory that ends at a special marker byte called '\0'; there is no stored length , so every string function walks forward until it finds that marker. Everything about safe copying is one question: does my destination row have enough bytes to hold the copy AND its ending marker before the walk runs off the edge?
Before you can understand strncpy, snprintf, or strlcpy, you must be able to see what a C string actually is in memory. The parent note throws around char buf[8], '\0', sizeof, size, and "off the end" as if they were obvious. This page earns every one of those, in order, from nothing.
char
A byte is the smallest addressable chunk of memory — think of it as one tiny numbered box that holds one number (0–255). In C, the type char is one byte. A single letter like 'h' is stored as its number code (for 'h' that number is 104).
The picture you must hold in your head for the entire topic: memory is a long strip of numbered boxes , and each box holds one number that we can read as a letter.
Intuition Why boxes, not lettersThe computer never stores "letters". It stores numbers in boxes . Whether a box "is" the letter h or the number 104 is only a matter of how we choose to read it . This is why a stray byte can silently become garbage text — same box, different reading.
char buf[8]
This reserves exactly 8 consecutive boxes and gives the whole row the name buf. The individual boxes are buf[0], buf[1], …, buf[7]. Index counting starts at 0 , so the last legal box is buf[7], not buf[8].
Common mistake Off-by-one from the startWhy it feels wrong: the size is 8, so surely buf[8] exists? No. 8 boxes numbered from 0 are 0..7. buf[8] is the 9th box — it does not belong to buf. Touching it is exactly the "walk off the end" the whole topic is about.
Notice there is nothing built into buf that says "I am 8 long." The number 8 lives only in your source code , not in memory. This missing length is the reason C string functions are dangerous — see C strings and the null terminator .
Definition The null terminator'\0'
'\0' is a single char whose stored number is 0 (zero). It is not the digit '0' (whose number is 48). C uses this zero-byte as a signpost meaning "the string ends here."
This is the single most important symbol in the whole topic. A "string" in C is defined as:
Intuition Why a marker instead of a stored length?Because C strings have no length field . A function like strlen finds the length by starting at box 0 and walking right until it hits a '\0' . Remove the marker and the walk never stops — it reads box after box into unrelated memory. That runaway walk is the failure mode of strncpy in the parent note. Deep dive: sizeof vs strlen .
sizeof(buf)
sizeof(buf) is the total number of boxes reserved for the array — a compile-time fact about capacity. For char buf[8] it is 8 , whether or not the boxes are filled.
strlen(buf)
strlen(buf) counts ==visible characters before the '\0'== — it walks the row at run time . It does not count the '\0' itself.
Worked example Same buffer, two numberschar buf [ 8 ]; strcpy (buf, "hi" ); // fills 'h' 'i' '\0' // sizeof(buf) == 8 (capacity — never changes) // strlen(buf) == 2 (content — up to the marker) The 5 boxes after the marker still exist (capacity 8) but hold whatever junk they held before.
dest and src
Every copy function takes a destination (dest, where bytes are written) and a source (src, where bytes are read). Water flows src → dest. Overflow is always a dest problem: dest is too small for what src pours in.
size_t
The type of the "how big" argument. It is just an unsigned (non-negative) whole number used for sizes and counts. When you see size_t r = strlcpy(...), r is a plain count of characters — nothing exotic.
A function can hand back one number when it finishes. The parent note leans hard on this: snprintf and strlcpy return "how many characters I wanted ," which you compare against capacity to detect truncation.
Now the parent note's opening example reads itself:
char buf [ 8 ]; strcpy (buf, "hello world" ); // 12 chars → needs 13 boxes "hello world" needs 11 + 1 = 12 buf has only 8. strcpy doesn't know or check sizeof(buf), so it keeps writing past box 7 into boxes 8, 9, 10… that belong to other things — the essence of Buffer overflow and stack smashing and Memory safety and undefined behavior .
char buf 8 is a row of boxes
Null terminator marks the end
A C string is boxes plus marker
strlen must be at most sizeof minus 1
Overflow walks off the row
Safe alternatives strncpy snprintf strlcpy
How many boxes does char buf[8] reserve, and what is the last legal index? 8 boxes; the last legal index is buf[7] (indexing starts at 0).
What number is stored in the '\0' byte, and how is it different from '0'? '\0' stores 0; '0' stores 48 — they are completely different bytes.
Why does storing k visible characters require k+1 boxes? One extra box is needed for the '\0' end-marker.
What does sizeof(buf) measure versus strlen(buf)? sizeof = total capacity in bytes (compile-time); strlen = visible characters up to the '\0' (walked at run time).
State the safety inequality relating content to capacity. strlen(buf) <= sizeof(buf) - 1.
How does a string function "know" where a string ends if C stores no length? It walks forward from box 0 until it finds a '\0' byte.
In copy(dest, src, size), which side can overflow and why? dest — it may be too small to hold what src pours in.
Give the one-line rule that detects truncation. Truncation happened iff return >= size.
Parent: Safe alternatives C strings and the null terminator sizeof vs strlen strcpy and sprintf (unsafe) Buffer overflow and stack smashing Memory safety and undefined behavior Format strings and printf family 
