What you are doing is undefined behavior. The compiler is allowed to assume that you will never use more than sizeof int64_t for the variable member int64_t c. So if you try to write more than sizeof int64_t(aka sizeof c) on c, you will have an out-of-bounds problem in your code. This is the case … Read more
strcpy is a unsafe function. When you try to copy a string using strcpy() to a buffer which is not large enough to contain it, it will cause a buffer overflow. strcpy_s() is a security enhanced version of strcpy(). With strcpy_s you can specify the size of the destination buffer to avoid buffer overflows during … Read more
Arrays in C are non-assignable and non-copy-initializable. That’s just how arrays are in C. Historically, in value context (on the RHS of assignment) arrays decay to pointers, which is what formally prevents assignment and copy-initialization. This applies to all arrays, not only to char arrays. C language inherits this arrays behavior from its predecessors – … Read more
as Evan pointed out, it is possible to do something like char* s = strcpy(malloc(10), “test”); e.g. assign malloc()ed memory a value, without using helper variable. (this example isn’t the best one, it will crash on out of memory conditions, but the idea is obvious)
The local variables have a lifetime which extends only inside the block in which it is defined. The moment the control goes outside the block in which the local variable is defined, the storage for the variable is no more allocated (not guaranteed). Therefore, using the memory address of the variable outside the lifetime area … Read more
C does not and never has had a native string type. By convention, the language uses arrays of char terminated with a null char, i.e., with ‘\0′. Functions and macros in the language’s standard libraries provide support for the null-terminated character arrays, e.g., strlen iterates over an array of char until it encounters a ‘\0’ … Read more
It depends on what you mean by “empty”. If you just want a zero-length string, then your example will work. This will also work: buffer[0] = ‘\0’; If you want to zero the entire contents of the string, you can do it this way: memset(buffer,0,strlen(buffer)); but this will only work for zeroing up to the … Read more
The first example doesn’t work because you can’t assign values to arrays – arrays work (sort of) like const pointers in this respect. What you can do though is copy a new value into the array: strcpy(p.name, “Jane”); Char arrays are fine to use if you know the maximum size of the string in advance, … Read more
strcpy(ptr2, ptr1) is equivalent to while(*ptr2++ = *ptr1++) where as strdup is equivalent to ptr2 = malloc(strlen(ptr1)+1); strcpy(ptr2,ptr1); (memcpy version might be more efficient) So if you want the string which you have copied to be used in another function (as it is created in heap section) you can use strdup, else strcpy is enough.
what could be done to see this effect Compile and run this code: void dump5(char *str); int main() { char s[5]={‘s’,’a’,’\0′,’c’,’h’}; char membuff[5]; char strbuff[5]; memset(membuff, 0, 5); // init both buffers to nulls memset(strbuff, 0, 5); strcpy(strbuff,s); memcpy(membuff,s,5); dump5(membuff); // show what happened dump5(strbuff); return 0; } void dump5(char *str) { char *p = … Read more