Issue using displaying Image with NSMutableArray - ios

Question

I am very new to xcode and I am playing around with image arrays and NSMutableArray. When I run the code below it just print out the names of the image for example "image1.png". Any tips to fix this problem would be appreciated. Thanks.
- (NSMutableArray*)restaurantArray;
{
if (_restaurantArray == nil) {
_restaurantArray = [[NSMutableArray alloc] initWithObjects:
#"image1.png",
#"image2.png",
#"image3.png",
nil];
}
return _restaurantArray;
}
-(NSString*) randomRestaurant
{
int random = arc4random_uniform(self.restaurantArray.count);
return [self.restaurantArray objectAtIndex:random];
}

Answer 1

You could do it in this way also.
- (NSMutableArray*)restaurantArray;
{
if (_restaurantArray == nil) {
_restaurantArray = [[NSMutableArray alloc] initWithObjects:[UIImage imageNamed:#"image1.png"],
[UIImage imageNamed:#"image2.png"],
[UIImage imageNamed:#"image3.png"],
nil];
}
return _restaurantArray;
}

Answer 2

Do you mean this?:
NSString *imageName = [self randomRestaurant];
UIImage *image = [UIImage imageName:imageName];

Answer 3

If you want to display the image then do this:
Suppose you have an imageview name imgView,
imgView.image = [UIImage imageNamed:[self randomRestaurant]];
Or, if you just want the image, then call this:
UIImage *img = [UIImage imageNamed:[self randomRestaurant]];
In you given code, you are just returning the name of the image, not the actual image. If you want an array of image then you can do it by this way:
-(NSMutableArray*) restaurantArray;
{
if (_restaurantArray == nil) {
UIImage *img = [UIImage imageNamed:#"image1.png"];
UIImage *img1 = [UIImage imageNamed:#"image2.png"];
UIImage *img2 = [UIImage imageNamed:#"image3.png"];
_restaurantArray = [[NSMutableArray alloc] initWithObjects:
img,
img1,
img2,
nil];
}
return _restaurantArray;
}
-(UIImage*) randomRestaurant
{
int random = arc4random_uniform(self.restaurantArray.count);
return [self.restaurantArray objectAtIndex:random];
}
Hope this helps.. :)