调用printf（）时为EXC_BAD_ACCESS

Question

What's wrong with this?

int main(int argc, char** argv) {

printf("%% ");
size_t len;
ssize_t read;
char* line;
int size;

read = getline(&line, &len,stdin);
printf("my name: %s\n",argv[0]);
printf("%s",line);
char* args[]= {"yoyoyo","hi","me",NULL};

return 0;
}

Debugging shows Exception: EXC_BAD_ACCESS (code=1, address=0xa66647360)) on the

printf("my name: %s\\n",argv[0]); line.

Answer 1

You're forgetting to initialize the values supplied to getline() .

Try with char *line = NULL; and size_t len = 0; instead.

The man 3 getline man page has an example you could adapt.

Answer 2

I suspect your problem is not with that line but with this:

read = getline(&line, &len,stdin);

line is declared as a char * and isn't pointed to anything. Thus, you're reading in data to the memory location of an uninitialized pointer, which is why you're getting an access error.

Try either statically allocating line:

char line[256]; // or whatever line length you want
read = getline(&line, &len, stdin);

or using malloc:

char *line = malloc(sizeof(char) * 256);
read = getline(line, &len, stdin);

but be careful of overflowing whatever length you set. Good luck!