Intricacies of strcpy_s in C++

Question

I am having a difficult time obtaining the correct size of a string in order to satisfy strcpy_s. For example if I specify

char buffer = {0};
char *str1 = (char*)&buffer;    
strcpy_s(str1,sizeof("This is a string\n"),"This is a string\n");

Then it will work as expected. If however I declare the following:

char buffer = {0};
char *str1 = (char*)&buffer;
const char* string1 = "This is a string.....";
strcpy_s(str1, ?????,string1);

If I use anything other than a literal in place of ????? it will fail with a memory exception, for example if I use std:strlen(str1), etc. Any size literal for ???? will work. Of course using a fixed literal is not acceptable.

This is a major re-edit of the original question and I apologise to the people who have answered to date. However none of the the answers below have worked.

The second argument should be the size of the destination, that is the size of str1. In other words, how much space is available. — zdf, Dec 10 '15 at 14:38
[Learn the difference between arrays and pointers](https://stackoverflow.com/questions/3959705/arrays-are-pointers). — Simple, Dec 10 '15 at 14:40
Your first attempt is not correct, even if it “works” (at some point it will crash). It is not correct for you have allocated a buffer that can hold **one character, only** and you are copying more than one character. The correct call would be `strcpy_s(str1, sizeof(buffer),"This is a string\n");`. However, you will not get the expected result since the buffer is only 1 character long. To see something you should allocate a large enough buffer: `char buffer[128]`, for instance. This buffer will hold a 127 characters long string (plus terminating null). — zdf, Dec 10 '15 at 15:51

Praveen · Accepted Answer · 2015-12-10T14:43:56.140

2

"This is a string" is a character array. When you say sizeof(Array)/sizeof(type) it will give the size of the array

When you define the string as const char* then the sizeof(pointer) gives the size allocated for the pointer no the array size

const char* ptr = "This is a string\n";
std::cout<<sizeof("This is a string\n")<<std::endl; //==>18
std::cout<<sizeof(ptr)<<std::endl; //==>4

edited Dec 10 '15 at 14:43

answered Dec 10 '15 at 14:38

Praveen

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score 2 · Answer 2 · answered Dec 10 '15 at 14:40

First of all, the second parameter is the size of the destination buffer, not the size of the source buffer.

so the correct way is:

char str1[100];
strcpy_s(str1, sizeof str1, "Whatever string");

or

int n = 100;
char *str1 = new char[n];
strcpy_s(str1, n, "whatever string");

For an array (first example) sizeof returns the size of the array.

For a pointer (second example) sizeof returns the size of the pointer (which is not what you want)

Aiden Deom · Answer 3 · 2015-12-11T14:17:49.943

In your second example, string1 is of type const char*. sizeof will return the size of the pointer, rather than the length of the string literal you are pointing to.

The first example works because a string literal is a const char[], and sizeof will correctly return the length of the string (but with the null terminating character as well). It's only coincidental that this works because char is 1 byte. Do not use sizeof to get string lengths.

To make your second example work, try using std::strlen.

Intricacies of strcpy_s in C++

3 Answers3