Do not understand about strcpy in C++

char* s1; strcpy(s1,"smilehihi"); s1[6] = 'a'; 

When I compile, VS do not have any errors. But in the runtime, my code makes mistake. I think I do not really understand about strcpy

4 Answers

The main issue here is not the strcpy() function but the fact that you don't allocate any memory for the string itself. If I were you, I would do something like

char* s1=(char*)malloc(SIZE); // the SIZE is the predefined maximum size of your string strcpy(s1,"smilehihi"); s1[6] = 'a'; 

Edit:

Just as an advice, consider using stpncpy(). It helps to avoid buffer overflow, and, in your case, will help you avoid exceeding the maximum size of char*

char * stpncpy(char * dst, const char * src, size_t len); 

The problem is that you have not allocated any space for what you wish to store in s1: "smilehihi". You declare s1 as a pointer variable, but it needs something to point at. You can allocate space by using the new operator.

char* s1 = new char[stringLength + 1]; //stringLength = length of string stored // + 1 to hold null terminator character strcpy(s1, "smilehihi"); s1[6] = 'a'; 

You have to declare #define _CRT_SECURE_NO_WARNINGS at the top of your main file to avoid an error during compilation due to strcpy() being deprecated.

You need to allocate variable first by malloc() or by using the keyword new . Also deallocate the memory at the end

At first you should allocate char* s1.

char *s1 = new char[9]; // C++ version 

or you can you use C version:

char *s1 = (char*)malloc(9); 

Then you can use following code:

strcpy(s1, "smilehihi"); s1[6] = 'a';