Casting char array into int array using atoi

Question

Hello I'm attempting to convert a char into an int. I have a char array that was inputted through scanf, "10101" and I want to set an int array's elements equal to that char arrays elements.

example input:

10101

char aBuff[11] = {'\0'};
int aDork[5] = {0};

scanf("%s", aBuff); //aBuff is not equal to 10101, printing aBuff[0] = 1, aBuff[1] = 0 and so on

Now I want aDork[0] to equal aBuff[0] which would be 1.
Below is what I have so far.

//seems to not be working here
//I want aDork[0] to  = aBuff[0] which would be 1
//But aDork[0] is = 10101 which is the entire string of aBuff   
//aBuff is a char array that equals 10101
//aDork is an int array set with all elements set to 0

int aDork[5] = {0}
printf("aBuff[0] = %c\n", aBuff[0]); //value is 1
aDork[0] = atoi(&aBuff[0]); //why doesnt this print 1? Currently prints 10101
printf("aDork[0] = %d\n", aDork[0]); //this prints 1

printf("aBuff[1] = %c\n", aBuff[1]); //this prints 0
printf("aBuff[2] = %c\n", aBuff[2]); //this prints 1

Casting is (forced) type-conversion. Not parsing, like atoi. — Deduplicator
– Deduplicator, Commented Dec 2, 2014 at 22:32
so if I want each element of the char array to become a literal int 1 or int 0 how can I achieve this? — Yawn
– Yawn, Commented Dec 2, 2014 at 22:34
the difference between a char number and a int number (of one digit) is 0x30 ('0') so: aDork[0] = aBuff[0]-'0'; aDork[1] = aBuff[1]-'0'; etc — user3629249
– user3629249, Commented Dec 3, 2014 at 3:54

Paul Roub · Accepted Answer · 2014-12-02 22:33:24Z

3

You ask:

aDork[0] = atoi(&aBuff[0]); // why doesnt this print 1? Currently prints 10101

It does so because:

&aBuff[0] == aBuff;

they're equivalent. The address of the first element in the array is the same address you get when referencing the array itself. So you're saying:

aDork[0] = atoi(aBuff);

which takes the whole string at aBuff and evaluates its integer value. If you want to get the value of a digit, do that:

aDork[0] = aBuff[0] - '0';   // '1' - '0' == 1, '0' - '0' == 0, etc.

and now

aDork[0] == 1;

Working example: https://ideone.com/3Vl3aI

answered Dec 2, 2014 at 22:33

Paul Roub

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2 Comments

Yawn Over a year ago

wouldn't I have to know what aBuff[1] is equal to? For example, if I want aDork[1] = aBuff[1] and I don't know if aBuff[1] is equal to '1' or '0', I wouldn't just be able to write aDork[1] = aBuff[1] - '0' because this would produce 1?

Paul Roub Over a year ago

No. If aBuff[1] == '0', then aBuff[1] - '0' is the same as '0' - '0', which is of course 0. See the example - it converts all the digits to integers in the aBuff array.

Ayman Khamouma · Accepted Answer · 2014-12-02 22:40:30Z

1

this code is working as expected, but you are not understanding how atoi works:

Now I want aDork[0] to equal aBuff[0] which would be 1

but

aDork[0] = atoi(aBuff);

means aDork[0] will store the integer value of aBuff. Meaning the value 10101 and not the string "10101"

PS: you didn't need an array of char for aDork:

int aDork = 0;
aDork = atoi(aBuff);

is enough.

answered Dec 2, 2014 at 22:40

Ayman Khamouma

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Comments

Jonathan Wood · Accepted Answer · 2014-12-02 22:44:15Z

1

Assuming aBuff contains a string of zeros and ones (that does not exceed the length of aDork), the following would transfer these values to an integer array.

for (int i = 0; i < strlen(aBuff); i++)
{
    aDork[i] = (aBuff[i] - '0');
}

answered Dec 2, 2014 at 22:44

Jonathan Wood

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