unsigned long and atoi in C

Question

Code:

char temp[] = "1010000001010011";
printf("%s\n", temp);
unsigned long num = atoi(temp);
printf("%lu\n", num);

Output:

1010000001010011
4081617243

Why is num not 41043?

`atoi` is for decimal numbers not binary. Use `strtol` for binary. — kaylum, Mar 24 '17 at 21:06
`atoi()` interprets its input as a string representation of a *decimal* number. If you want to be convert a number expressed in a different base -- say base 2 -- then use `strtol()`. — John Bollinger, Mar 24 '17 at 21:09
Note also that when interpreted as a decimal number, your input string very likely exceeds your system's largest representable `long` value. — John Bollinger, Mar 24 '17 at 21:11
"Why is num not 41043?" What's 41043 and what does it have to do with anything? — AnT stands with Russia, Mar 24 '17 at 21:22
@AnT isn't is completely obvious that `41043` is the decimal value of binary `1010000001010011`? — Weather Vane, Mar 24 '17 at 21:30
@Weather Vane: Firstly, no, it is not obvious. Secondly, no, I did not downvote. — AnT stands with Russia, Mar 24 '17 at 21:34
@AnT what is more obvious is that OP wanted a conversion from binary, and it's a small step to a calculator. — Weather Vane, Mar 24 '17 at 21:37
Thanks for the help...seems I need to convert to decimal in the char* before atoi... — Jerry G, Mar 25 '17 at 02:48

score 2 · Answer 1 · answered Mar 25 '17 at 03:13

atoi("1010000001010011"); attempts to convert the text as a sequence of decimal characters to an int. If int was wide enough, it would take on the decimal value of 1,010,000,001,010,011. Not surprisingly, that value is outside OP's int range, so the result is undefined behavior or UB.

To convert the string as if it was binary characters, use strtol() or strtoul()

char temp[] = "1010000001010011";
char *endptr;
int base = 2;
unsigned long ulnum = strtoul(temp, &endptr, base);
if (endptr == temp) puts("No conversion");
else printf("%lu\n", ulnum);  // 41043

score 1 · Answer 2 · edited May 23 '17 at 12:26

1

atoi returns an int value. The decimal number 1010000001010011 which is hex 39696F348895B does not fit a 32-bit variable and was truncated to hex F348895B which is the signed decimal value -213350053. However it is undefined behaviour to assign that value to unsigned long, although the compiler produced code that does assign that bit pattern, which as unsigned long happened to be 4081617243.

Many thanks to @AnT and to @chux

edited May 23 '17 at 12:26

Community

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answered Mar 24 '17 at 21:15

Weather Vane

33,872
7
36
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2

Overflow in `atoi` results in *undefined behavior*, not in truncation. – AnT stands with Russia Mar 24 '17 at 21:23

score -2 · Answer 3 · answered Mar 24 '17 at 22:12

-2

Beacouse atoi() converts the string into an integer, does not convert from binary. For example the code

char str[]="123";
printf("%d",atoi(str));

will print 123 as an integer

answered Mar 24 '17 at 22:12

AlessandroF

542
5
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unsigned long and atoi in C

3 Answers3