I'm new to C, coming from Java.
I made the following trivial program that compiles fine, but nano throws me a Segmentation Fault whenever I run it. The point of it is to traverse through the array and have it print out each element on a separate line.
int main()
{
int array[5] = {1, 2, 3, 4, 5};
int i = 0;
for (i = 0; i < sizeof(array); i++)
{
puts(array[i]);
}
}
What am I doing wrong?
First, puts takes an null terminated strings, not an integer. Second, to determine the number of elements in the array, use sizeof(array)/sizeof(array[0]), because sizeof(array) is the total number of bytes of the array. Third, use int main(void) for standard C. Try this:
int main(void)
{
int array[5] = {1, 2, 3, 4, 5};
int i = 0;
for (i = 0; i < sizeof(array) / sizeof(array[0]); i++)
{
printf("%d\n", array[i]);
}
}
return 0, and its always a good idea to put brackets around operations on either side of a comparison since Strange Things have been known to happen when they're forgotten. +1 Though.
main(), it is equivalent to return 0;. I don't like this — but it agrees with what C++98 decided (and I don't like that decision either), so that isn't wholly fair. I don't see any relevant problems with missing parentheses, either — it is not unreasonable to expect people to be vaguely familiar with the table of operator precedences, and there are no bitwise operations here to confuse things.return 0 bit though, its probably just because I've been forced to use C89 all my life, all this new flim-flammery is quite odd!
sizeofgives the size of the object, in bytes. That is not the same as the number of elements in an array (unless the size of an element is 1 byte). To get the number of elements, you must divide the value by the size of a single array element. Also, you can't pass aninttoputs, which expects a string, i.e. achar *.gcc -std=c11 -pedantic-errors.