strange return behavior in strcat()

Question

void mystrcat(char* to, const char* from) {
    while (*to) to++;
    while (*from) *to++ = *from++;
    *to = '\0';
}

int main() {
    char addthis[]= "rest of the sentence";
    char start_of[] = "going to add ";


    mystrcat(start_of, addthis);

    cout << "after strcat(): " << start_of<< endl;
}

even if i replace the function mystrcat to follows, behaviour is same.

char* mystrcat(char* to, const char* from) {
    while (*to) to++;
    while (*from) *to++ = *from++;
    *to = '\0';
    return to;
}

strange for me, when i call mystrcat i dont assign to a char* still no compiler's complain. what am i missing here? follow up can u optimize my code with void return type if anyway

You don't have any room in the first string to concatenate into. And this is not C. — Eugene Sh., Jul 22 '20 at 18:22
You are appending to `start_of[]` which does not have space allocated for the extra data. Undefined behavior. — Michael Burr, Jul 22 '20 at 18:22
what if i dont want to hardcode any specific value and keep it open for as much space it can take. is there a way? — EagleEye, Jul 22 '20 at 18:35

score 2 · Answer 1 · answered Jul 22 '20 at 18:22

2

The string start_of is declared to be only long enough to hold the string it is initialized with. So attempting to append to it writes past the end of the array. This invokes undefined behavior.

You need to make the array large enough to hold the concatenated string.

char start_of[50] = "going to add ";

answered Jul 22 '20 at 18:22

dbush

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thanks. but what if i dont want to make a hardcoded value like 50? – EagleEye Jul 22 '20 at 18:31
@EagleEye: If you don't want a hardcoded value, then you can't store it in a local array (unless you want to use platform-specific extensions such as [VLAs](https://en.wikipedia.org/wiki/Variable-length_array)). You must use dynamic memory allocation instead, for example [`std::malloc`](https://en.cppreference.com/w/cpp/memory/c/malloc) and [`std:realloc`](https://en.cppreference.com/w/cpp/memory/c/realloc) for C-style memory allocation or [new](https://en.cppreference.com/w/cpp/language/new) for C++-style memory allocation. You can also use some kind of C++ container, such as `std::string`. – Andreas Wenzel Jul 22 '20 at 18:39

Abhishek Bhagate · Answer 2 · 2020-07-22T18:39:10.610

You don't need to always assign the return value to some variable if you are returning from a function in C. The other functions like printf scanf also returns value but they need not give any error if you are not assigning them to some variable while calling.

Also, as a side note, your mystrcat function is running on undefined behavior. You are passing it two char array and appending to the first char array itself when it doesn't have any further space allocated to it. So, you should change it.

You should declare the char array of first argument that you are passing to the function to be long enough to be able to hold the data after concatenation. You can change your program as follows -

void mystrcat(char* to, const char* from) {
    while (*to) to++;
    while (*from) *to++ = *from++;
    *to = '\0';
}

int main() {
    char addthis[]= "rest of the sentence";
    char start_of[100] = "going to add ";


    mystrcat(start_of, addthis);

    cout << "after strcat(): " << start_of<< endl;
}

as i commented above, what if i dont want to hardcode '100' and keep it open for as much space it can take. is there a way? — EagleEye, Jul 22 '20 at 18:34

strange return behavior in strcat()

2 Answers2