CODEWITHSUNDEEP
csegmentation-fault
Cesar
Cesar

Reputation: 115

Printing the value of a char array

I'm trying to write a function that takes in a const char* and iterates through it until it matches the first character of the const char* hello. It then prints the amount of characters it had to advance and the mth character. As of now, I have gotten it to point to where I want to, but I can't seem to access it

const char* hello = "hello";
const char* no_bears = "ERROR:  Message must contain 'THIS IS BEAR TERRITORY!' yeeee";

int is_commit_msg_ok(const char* msg) {
  /* COMPLETE THE REST */
  int m = 0;
  char* message = &msg[0];
  while (message[m] != hello[0]) {
    if (message[m] == '\0') {
        return 0;
    }
    m++;

  }
  printf("%d\n",m );
  printf("%s\n", message[m]); \\ causes segmentation fault

When I do message[m], however, I get a segementation fault: 11 error, and I don't really know how to fix this...

Upvotes: 0

Views: 3660

Answers (2)

user3368555
user3368555

Reputation:

You can use printf("%c\n", message[m]); because you printing just single char not a string with "%s".

Upvotes: 1

rodrigo
rodrigo

Reputation: 98348

In this line:

printf("%s\n", message[m]);

message is of type char* so message[m] is of type char but %s expects a char*.

Passing the wrong type to printf() is undefined behaviour. But what it probably does is to interpret the char value as a pointer and try to read the memory there. Since it is not a valid pointer, it segmentation-faults.

Solution, use %c, that prints a char.

printf("%c\n", message[m]);

Many modern compilers will warn you about these kind of errors if you enable the warnings. For example in GCC/Clang, you should really use -Wall.

Upvotes: 3

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