Why does "strcat" on an array cause overwriting?

Question

strcat(argv[1], ".txt");
puts(argv[1]);
strcat(argv[2], ".txt");
puts(argv[2]);
puts(argv[1]);

Say the first([1]) argument is called "a" and the second([2]) "b". This code gives the following output: a.txt txt.txt a.txt.txt

Why is this happening?? It seems as if the strcat starts overwriting arg[2]?

"string management" in C is different. – alk Oct 23 '17 at 08:49 — alk, Oct 23 '17 at 08:49

AnT stands with Russia · Answer 1 · 2017-10-23T06:36:13.743

2

You are not supposed to make any assumptions about amount of memory allocated to store strings pointed by argv[i]. In particular, you are not allowed to assume that you can expand these strings by strcat-ing anything to them. Doing so will generally lead to undefined behavior. This is exactly what you observed in your experiment.

While it is legal to modify the strings pointed by argv[i] without changing their length, it is not generally possible to "expand" them in place. It is conceivable that these strings might immediately follow each other in memory, meaning that writing anything past the end of string pointed by argv[1] would overwrite string pointed by argv[2] and so on.

edited Oct 23 '17 at 06:36

answered Oct 23 '17 at 06:28

AnT stands with Russia

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I see, so how would I append the .txt ending? – Hedhed Oct 23 '17 at 06:29
@Hedhed make another string. Allocate enough bytes with (malloc) – Antti Haapala -- Слава Україні Oct 23 '17 at 06:36
@Hedhed: What is your intended effect of appending `.txt` directly to string stored at memory location pointed by `argv[1]` instead of making a completely independent string? – AnT stands with Russia Oct 23 '17 at 06:38

Why does "strcat" on an array cause overwriting?

1 Answers1