convert string into signed int

Question

I want to convert a string into a signed int. Following is the requirement. I have stored hex value as a string in buffer. Now I want to convert that value into signed int.

buf = "fb869e" Convert this into signed int. So o/p should be -293218. but when I'm trying to convert using strtol I'm getting 16483998. So what I should I do?

Where did you get that 0xfb869e is -293218? strtol is actually right. — vanza, Sep 25 '10 at 05:34
why output should be -293218? strtol return type is long int, btw. — Nyan, Sep 25 '10 at 05:36
possible duplicate of [C++ convert hex string to signed integer](http://stackoverflow.com/questions/1070497/c-convert-hex-string-to-signed-integer) — Eli Bendersky, Sep 25 '10 at 05:46
Kind of a duplicate: http://stackoverflow.com/questions/1070497/c-convert-hex-string-to-signed-integer — Adi_aks, Sep 25 '10 at 05:36

score 5 · Answer 1 · answered Sep 25 '10 at 06:16

The hexadecimal number 0xfb869e is not negative. The inbuilt number conversion functions will not convert it to a negative value, since its value is positive.

What you are saying is that this is the unsigned hexadecimal equivalent of a 24-bit 2s complement negative number, and you want that number. The way to get that is to convert it to the positive number, then use calculations to convert it to the 24-bit 2s complement equivalent:

char *buf = "fb869e";
long n;

n = strtol(buf, NULL, 16);
if (n > 0x7fffffL)
    n -= 0x1000000L;

score 1 · Answer 2 · answered Sep 25 '10 at 06:57

1

Others have suggested strtol(). I just want to mention sscanf() as an alternative, eg:

int i;
char *buf = "fb869e";
if (sscanf(buf, "%x", &i) == 1)
   ...

answered Sep 25 '10 at 06:57

Remy Lebeau

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score 0 · Answer 3 · answered Jun 20 '12 at 23:39

0

Use strtol which converts string to a long integer.

answered Jun 20 '12 at 23:39

Alex Lockwood

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Marcelo Cantos · Answer 4 · 2010-09-25T06:33:27.693

0

0xfb869e == 0x00fb869e == 16483998

As a signed integer, the high bit must be set to produce a negative number. Since the high bit of the given number isn't set, it must be positive.

If you want the number to be treated as a 24-bit number, you'll have to pad bit 23 out to the remaining high-bits. Here's one way to do it:

long n = strtol(...);
if (n > 0xffffff) n |= 0xff000000;

edited Sep 25 '10 at 06:33

answered Sep 25 '10 at 05:34

Marcelo Cantos

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char buf = "fb869e" - I want to convert into signed int. The o/p should be -293218. I just tried converting string into long int. But I'm not getting the negative value. – ulaganathan Sep 25 '10 at 05:51
char buf = "fb869e". I want to convert this into signed int. So o/p should be -293218. I just tried converting string into long using strtol. But I'm not getting the negative values. My hex values are follows. I need o/p in signed int. "C programming" 80000e fcc15e 7ffffe 0b778e 80000e ee4dae 7ffffe 1b1e5e 80000e de2b9e 7ffffe 2b8c4e 80000e ce2e1e 7ffffe – ulaganathan Sep 25 '10 at 05:58
@ulganathan: There was a bug in my answer. Use `if (n > 0x7fffff)...` instead. – Marcelo Cantos Sep 25 '10 at 06:38

score 0 · Answer 5 · answered Sep 25 '10 at 05:35

strtol converts string to long integer

http://www.cplusplus.com/reference/clibrary/cstdlib/strtol/

The output is correct and it will be 16483998

And if you use atoi, while it converts to string to integer, the correct value if out of the range of representable values, INT_MAX or INT_MIN is returned.

http://www.cplusplus.com/reference/clibrary/cstdlib/atoi/

score 0 · Answer 6 · answered Sep 25 '10 at 05:36

0

why should 0x00fb869e be a negative number? you should have to provide the base of your number system, to even be allowed to tell, whether a value in one format is a negative in another format

answered Sep 25 '10 at 05:36

Peter Miehle

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char buf = "fb869e". I want to convert this into signed int. So o/p should be -293218. My hex values are follows. I need o/p in signed int. 80000e fcc15e 7ffffe 0b778e 80000e ee4dae 7ffffe 1b1e5e 80000e de2b9e 7ffffe 2b8c4e 80000e ce2e1e 7ffffe – ulaganathan Sep 25 '10 at 05:57

score 0 · Answer 7 · answered Oct 16 '19 at 06:25

0

Try below block of code, its working for me.

char *p = "0x820";
uint16_t intVal;
sscanf(p, "%x", &intVal);

printf("value x: %x - %d", intVal, intVal);

Output is:

value x: 820 - 2080

answered Oct 16 '19 at 06:25

Dig The Code

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convert string into signed int

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