seg fault during printf on variable I just set
So I'm seg faulting when I call printf in the following situation. I just can't see what I'm doing wrong. Any ideas? Thanks a million. I've marked the spot in the code where i get the seg fault with a comment (in the first chunk of code).
...
char开发者_JAVA技巧* command_to_run;
if(is_command_built_in(exec_args, command_to_run)){
//run built in command
printf("command_to_run = %s\n", command_to_run); // <--- this is where the problem is
run_built_in_command(exec_args);
}
...
int is_command_built_in(char** args, char* matched_command){
char* built_in_commands[] = {"something", "quit", "hey"};
int size_of_commands_arr = 3;
int i;
//char* command_to_execute;
for(i = 0; i < size_of_commands_arr; i++){
int same = strcmp(args[0], built_in_commands[i]);
if(same == 0){
//they were the same
matched_command = built_in_commands[i];
return 1;
}
}
return 0;
}
You're passing the pointer matched_command
by value. Therefore it isn't changed by the call to is_command_built_in
. So it retains it's uninitialized value.
Try this:
char* command_to_run;
if(is_command_built_in(exec_args, &command_to_run)){ // Changed this line.
//run built in command
printf("command_to_run = %s\n", command_to_run); // <--- this is where the problem is
run_built_in_command(exec_args);
}
int is_command_built_in(char** args, char** matched_command){ // Changed this line.
char* built_in_commands[] = {"something", "quit", "hey"};
int size_of_commands_arr = 3;
int i;
//char* command_to_execute;
for(i = 0; i < size_of_commands_arr; i++){
int same = strcmp(args[0], built_in_commands[i]);
if(same == 0){
//they were the same
*matched_command = built_in_commands[i]; // And changed this line.
return 1;
}
}
return 0;
}
command_to_run
is uninitialized. The call to is_command_built_in
could as easily have crashed. Such is the nature of undefined behavior.
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