Segmentation Fault in C - c

Question

My code is giving me a segmentation fault and I can't seem to find what I'm doing wrong:
#include <stdio.h>
#include <string.h>
char find(char name[], char allNames[][10], int length)
{
int i=0;
for (i = 0; i < length; i++) {
if (strcmp(allNames[i],name) == 1) {
printf("%i",i);
return *name;
}
}
return -1;
}
main(){
char allNames[][10] = {"cat","dog","frog","log","bog"};
char name[] = "log";
int length=5;
printf("%s",find(name,allNames,length));
}
I'm really keen to understand all the mechanisms happening here and what I'm doing wrong for tomorrows exam. Thanks for your help!
EDIT: Really Appreciate the answers and information guys! I'm really quite new to C and just getting used to what every thing means. The particular exam question I am looking at is :
(a) The following function is intended to find the string name in the array
allNames. If found, it returns the position of name in the array. If not
found, it returns -1. Modify the code so that it works correctly.
int find(char name[], char allNames[][10])
{
for (i = 0; i < 10; i++) {
if (allNames[i] == name) {
return name;
}
}
return -1;
}
And I'm trying to get a program to work within these parameters. Cheers :)

Answer 1

http://coliru.stacked-crooked.com/a/d400c9a56d732446
#include <stdio.h>
#include <string.h>
char* find(char name[], char allNames[][10], int length)
{
int i=0;
for (i = 0; i < length; i++) {
if (!strcmp(allNames[i],name)) {
printf("%i",i);
return name;
}
}
return NULL;
}
int main(){
char allNames[][10] = {"cat","dog","frog","log","bog"};
char name[] = "log";
int length=5;
printf("%s",find(name,allNames,length));
}
Returning a single char will do you no good if you're trying to return a string. I would also suggest that you return a NULL if you cannot find the string.
Also, include the int before main; this is better style.

Answer 2

The direct reason for your Segmentation Fault here is because the code tried to print the char type with %s(which needs an address value).
void main()
{
char c = 'a';
printf("%s", c); // will cause Segmentation fault here
}
Back to your code, that is
char find(char name[], char allNames[][10], int length)//return char
printf("%s",find(name,allNames,length));
The minimal change to make it work as follows,
1) To return char*
char* find(char name[], char allNames[][10], int length)//return char*
{
int i=0;
for (i = 0; i < length; i++) {
if (strcmp(allNames[i],name) == 0) { // here should 0
printf("%i",i);
return name; // change name* to name
}
}
return NULL; // change to NULL
}
//to print
printf("%s",find(name,allNames,length));
2) to return position value
int find(char name[], char allNames[][10])
{
for (i = 0; i < 10; i++) {
if (allNames[i] == name) {
return i; // here, change to return i
}
}
return -1;
}
//then, you can print like this
printf("find at position: %d",find(name,allNames,length));
//or to print string by
int pos = find(name,allNames,length);
if(pos >= 0)
printf("find the string: %s",allNames[pos]);

Answer 3

This code is wrong on several levels.
gcc -Wall -Wextra reveals:
meh.c:15:1: warning: return type defaults to ‘int’ [-Wreturn-type]
main(){
^
meh.c: In function ‘main’:
meh.c:19:3: warning: format ‘%s’ expects argument of type ‘char *’, but argument 2 has type ‘int’ [-Wformat=]
printf("%s",find(name,allNames,length));
^
meh.c:21:1: warning: control reaches end of non-void function [-Wreturn-type]
}
^
What's up with that? Do you compile with warnings enabled?
I am ignoring the lack of indentation.
#include <stdio.h>
#include <string.h>
char find(char name[], char allNames[][10], int length)
What? How about: char *name, **list, int size)
{
int i=0;
Why set it to 0 here?
for (i = 0; i < length; i++) {
if (strcmp(allNames[i],name) == 1) {
printf("%i",i);
return *name;
Have you read strcmp's manpage? It returns ZERO when a string matches, so this code makes no sense.
*name is of type char, but you don't want to return a char. You want to return a pointer, no?
}
}
return -1;
Well, given that you feed that into %s in printf, what do you expect to hapen here? Normally one would return NULL.
}
main(){
This is obsolete syntax, I don't know where you stole it from. Use 'int main(void)'.
char allNames[][10] = {"cat","dog","frog","log","bog"};
Normally people just return such arrays with a NULL pointer, so that these can be iterated over and there is no need to pass anything about the size.
char name[] = "log";
Why not char *name = "log".
int length=5;
Incorrect. It hardcodes the amount of stored strings in allNames table.
printf("%s",find(name,allNames,length));
}