“Abort trap: 6” error in C?

问题

I'm a beginner to C but I have this code running on xcode through gcc on terminal:

#include <stdio.h>
#include <string.h> 
int main(){
    char name[12] = "Roman Mirov"; 
    printf("My name is %s\n", name);
    name[8] = 'k'; 
    printf("My name is %s\n", name);
    char greeting[] = "hello"; 
    printf("%s %s\n", greeting, name);
    strcpy(greeting, "greetings, "); 
    printf("%s%s\n", greeting, name);
    return 0;
}

And it outputs this:

My name is Roman Mirov
My name is Roman Mikov
hello Roman Mikov
Abort trap: 6

My question exactly is, why it generates error instead of showing the last line as output "greetings, Roman Mikov"?

回答1:

In this case, the destination greeting does not have enough space to contain the whole contents of source, so it is an out of bounds access which invokes undefined behavior.

To elaborate, the size of array greeting is determined by the size of the supplied initializer,

char greeting[] = "hello";

in this case, "hello" which makes the size as 6, including the null-terminator.

Now, later you try to put a much bigger string into the memory,

strcpy(greeting, "greetings, ");

where, the source is of 12 bytes in size, whereas, the destination only contains 6. This causes the boundary overrun and the result, UB. The crash (or abort) is one of the possible side-effects of UB.

回答2:

In this line, you are allocating an array of 5+1 characters:

char greeting[] = "hello";

In this line, you are attempting to write 11+1 characters into that array:

strcpy(greeting, "greetings, ");

回答3:

In this case,greeting variable is array of char with size is 6 (Because hello and \0 ).

So when you use strcpy(greeting, "greetings, "); to copy "greetings, " into greeting is cannot. Because greeting not enough to contain array with 11+1 character

=> Has error abort trap : 6 here

来源：https://stackoverflow.com/questions/41340717/abort-trap-6-error-in-c

标签

arrays

strcpy