variable value change after strcpy function in C - c

Question

i noticed that variable value have been changed after strcpy function,i don't know how and why this is happening.
this is my code:
#include <stdio.h>
#include <string.h>
int main()
{
int b = true;
char ch[3];
strcpy(ch,"int");
printf("b value is:%d\n",b);
return 0;
}
I even use a temporary integer variable. it's very strange , in this case, the b value is correct after strcpy but the temp is changed to 0, note that it's just happened when assigning 1 value to b and temp variables.
this is the second code
#include <stdio.h>
#include <string.h>
int main()
{
int b = true;
int tmp=b;
char ch[3];
strcpy(ch,"int");
printf("b value is:%d\n",b);
return 0;
}

Answer 1

strcpy(ch,"int");
is equivalent to
ch[0] = 'i';
ch[1] = 'n';
ch[2] = 't';
ch[3] = 0;
Seeing as ch only has three elements (ch[0]..ch[2]), this invokes undefined behaviour.

Answer 2

Firstly, you need to use number (0,1) in your boolean variables. If you want to use boolean in C as you use them in c++, you have to include stdbool.h library.
Concerning strcpy(dest, src) function, it copies the src string to dest and return à pointer to the copied string.
In reality the size of "int" is not 3 but 4 bytes. String always ends with '\0' character.
In other words :
If you don't understand, try to analyze this code :
#include <stdio.h>
#include <string.h>
int main()
{
int b = 1;
char ch[4];
strcpy(ch,"int");
printf("b value is:%d\n",b);
return 0;
}

Answer 3

I used online compiler and selected "C++" and the error is not happening.
but when I selected C++17 I got this warning "
input
main.cpp:9:12: warning: ‘void __builtin_memcpy(void*, const void*, long unsigned int)’ writing 4 bytes into a region of size 3 ov
w=] "*
as #PaulMcKenzie mentioned this is buffer overflow case.