A question about how to enter a very long number.

[66650755]

I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

The following is my trial,but I find that the "a" is always 0.

#include<stdio. h>
#include<stdlib .h>

int main()
{

char string1[50];
int a;

scanf("%s",stri ng1);

a= atoi("string1") ;

printf("%d",a);

return 0;
}

[Ben Bacarisse]

66******@qq.com writes:

I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

I would not try. Your system is unlikely to have any integer type
that can hold the value so you have to do something else. Since this
looks like homework, I doubt the answer "use a bignum library" is
suitable. You have to come up with some way to represent these
numbers, and until you do, no one can say how to convert your string
into such a number. One solution is to leave it as a string.

<snip>
--
Ben.

[Keith Thompson]

66******@qq.com writes:

I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

[...]

scanf("%s",stri ng1);

a= atoi("string1") ;

[...]

One problem is that you need to understand the difference between
string1
and
"string1"

Fixing that isn't enough to solve your problem;
111111111111111 111111111111111 111111111111111 11111 would require a
164-bit signed integer type to represent it, and your implementation
almost certainly doesn't provide such a type.

Also, the atoi function is perhaps acceptable for a quick-and-dirty
program, but its behavior if you give it something it can't handle is
unpredictable.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Martin Ambuhl]

66******@qq.com wrote:

I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

The following is my trial,but I find that the "a" is always 0.

[hopeless code snipped]
Try this as a place to start. You will obviously want to modify it.

#include <stdio.h>
#include <limits.h>
#include <stdlib.h>
#include <errno.h>
#include <string.h>
#include <stddef.h>

int main()
{
unsigned long long /* or unsigned long if you don't have
this type */
the_number = 0;
unsigned limit = CHAR_BIT * sizeof the_number;
char buffer[limit + 2], *endp, *nl;
ptrdiff_t nchars;
printf("Please type a binary number of at most %u digits.\n"
"A character other than 0 or 1 will terminate the number.\n"
"Remember to end your imput with an end-of-line character.\n"
"Type it here: ", limit);
fflush(stdout);
errno = 0;
if (!fgets(buffer, sizeof buffer, stdin)) {
fprintf(stderr,
"Something untoward happened with your input.\n");
if (errno)
perror("errno was set because:");
fprintf(stderr, "I'm giving up.\n");
exit(EXIT_FAILU RE);
}
if ((nl = strchr(buffer, '\n')))
*nl = 0;
else
fprintf(stderr, "That was a very long line. The end-of-line\n"
"character never made it into the buffer,\n"
"The string we will be converting is:\n\"%s\"\n",
buffer);
the_number =
strtoull /* or strtoul, if needed */ (buffer, &endp, 2);
if (errno)
perror("errno was set:");
if (*endp) {
nchars = endp - buffer;
fprintf(stderr, "The string ended with %c (%d)\n"
"rather than an EOL character.\n", *endp, *endp);
*endp = 0;
printf("The string converted was %u chars, and was\n"
"\"%s\".\n" , (unsigned) nchars, buffer);
}
printf
("Your number is %llu (base 10), %#llo (base 8), %#llx"
" (base 16)\n",
the_number, the_number, the_number);
return 0;
}

[Martin Ambuhl]

Ben Bacarisse wrote:

66******@qq.com writes:

>I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(1111111 11111.....11111 111),because I want to use that number to
do some additions.How can I do that?

I would not try. Your system is unlikely to have any integer type
that can hold the value so you have to do something else.

This is an ubwarranted claim. The number of implementations without a
64 bit integral type is becoming vanishingly small.
And 66650755 needs only 50 of them.

[Martin Ambuhl]

66******@qq.com wrote:

I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

I should point out that your question does not tell us the base of the
representation. All other answers that I have seen assume base 10, and
their statements about probable impossibility are correct with that
assumption. The code I posted assumes base 2. Higher bases may quickly
run out of room.

[Nick Keighley]

On 30 Oct, 06:25, Martin Ambuhl <mamb...@earthl ink.netwrote:

Ben Bacarisse wrote:

[e-mail] writes:

I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

I would not try. *Your system is unlikely to have any integer type
that can hold the value so you have to do something else. *

This is an ubwarranted claim. *The number of implementations without a
64 bit integral type is becoming vanishingly small.
And 66650755 needs only 50 of them.

[Nick Keighley]

On 30 Oct, 06:25, Martin Ambuhl <mamb...@earthl ink.netwrote:

Ben Bacarisse wrote:

[e-mail] writes:

I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

I would not try. *Your system is unlikely to have any integer type
that can hold the value so you have to do something else. *

This is an ubwarranted claim. *The number of implementations without a
64 bit integral type is becoming vanishingly small.
And 66650755 needs only 50 of them.

I must be in need of coffee. I don't follow this. What is
the conection between a number that is approx 1e50 and your
666... number? Even 1...1 base 2 is larger than that.

--
Nick Keighley

[osmium]

<66******@qq.co mwrote:

>I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

The following is my trial,but I find that the "a" is always 0.

#include<stdio. h>
#include<stdlib .h>

int main()
{

char string1[50];
int a;

scanf("%s",stri ng1);

a= atoi("string1") ;

printf("%d",a);

return 0;
}

Start by getting your program to run with a more modestly sized number, say
three or four decimal digits. Once that works you can address the problem
of how to handle Really Big Numbers. The link below might be helpful in
attacking phase 2.

http://en.wikipedia.org/wiki/Binary_coded_decimal

[Mark McIntyre]

66******@qq.com wrote:

I would like to enter a string such as "111111111111.. ...11111111" (50
ones),and then convert it to a
number(11111111 1111.....111111 11),because I want to use that number to
do some additions.How can I do that?

The following is my trial,but I find that the "a" is always 0.

int a;

What is the largest number you can hold in an int?

scanf("%s",stri ng1);

don't use scanf - what happens if the user types
"supercalifragi listicexpialido coius" instead of a number?

a= atoi("string1") ;

don't use atoi -it doesn't do any error checking.

Also, why are you passing a string literal "string1" to atoi? You surely
want to pass a variable to it...

--
Mark McIntyre

CLC FAQ <http://c-faq.com/>
CLC readme: <http://www.ungerhu.com/jxh/clc.welcome.txt >