需要帮助 - 指针 [英] Need help - pointers
问题描述
你好，
我正在编写一个带十进制数字的程序并返回它的
二进制等价物。
问题是我不知道如何从（函数）返回
二进制值作为一个共同终止的字符数组。例如：我
想要使用4位转换数字3，即3< ----" 0011" ;.
执行后，我得到了这个警告：
警告：从不兼容的指针类型返回
行：49：警告：函数返回局部变量的地址
这是我的代码：
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int power（int base，int m）;
char * dec2bin（int input，int m）;
int main（）
{
int input = 0;
int m = 0;
printf（" Number to convert\\\
）;
scanf（"％d"，& input）;
printf（"位数\ n）;
scanf（"％d"& m）;
printf（"输入％d转换为％s \ n，输入，dec2bin（输入，m））;
退出（0）;
}
char * dec2bin（int input，in tm）
{
static int i = 0;
char * array [m + 1];
int limit = 0;
limit = power（2，m）;
const int MASK = limit / 2; //二进制等价物是：1 + m 0位
array [m] =''\ 0'';
for（i = 0 ; i< m; i ++）
{
if（input& MASK）{
strcpy（array [i]，" 1"）;}
else {
strcpy（array [i]，" 0"）;}
input = input< < 1;
}
返回（数组）;
}
int power（int base ，int m）
{
static int i;
int p = 1;
for（i = 1; i< = m; ++ i）
{
p = p * base;
}
返回p;
}
Hi There,
I''m writing a program that takes a decimal number and returns it''s
binary equivalent.
the problem is that I don''t know how to return (from function) the
binary value as an array of cocatenated characters. For example : I
want to convert number 3 using 4 bits, ie 3 <----"0011".
After executing, I''m getting this warning:
warning: return from incompatible pointer type
line:49: warning: function returns address of local variable
Here it is my code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int power (int base, int m);
char* dec2bin(int input, int m);
int main ()
{
int input = 0;
int m = 0;
printf("Number to convert\n");
scanf("%d", &input);
printf ("Number of bits\n");
scanf ("%d",&m);
printf("input %d is converted to %s\n",input,dec2bin(input, m));
exit (0);
}
char* dec2bin(int input, int m)
{
static int i = 0;
char* array[m+1];
int limit = 0;
limit = power(2, m);
const int MASK = limit/2; // the binary equivalent is: 1 + m 0 bits
array[m] = ''\0'';
for (i= 0; i < m ;i++)
{
if (input & MASK){
strcpy(array[i],"1");}
else{
strcpy(array[i],"0");}
input = input << 1;
}
return (array);
}
int power (int base, int m)
{
static int i;
int p=1;
for (i = 1; i <= m; ++i)
{
p =p * base;
}
return p;
}
推荐答案
Nezhate写道：
Nezhate wrote:
你好，
我正在编写一个带十进制数字的程序并返回它'
二进制等价物。
问题是我不知道如何从（函数）返回
二进制值作为一个共同终止的字符数组。例如：我
想要使用4位转换数字3，即3< ----" 0011" ;.
执行后，我得到了这个警告：
警告：从不兼容的指针类型返回
行：49：警告：函数返回局部变量的地址
这是我的代码：
#include< stdio.h>
#include< stdlib.h>
#include< string.h>
int power（int base，int m）;
char * dec2bin（int input，int m）;
int main（）
{
int input = 0;
int m = 0;
printf（" Number to convert\\\
）;
scanf（"％d"，& input）;
printf（"位数\ n）;
scanf（"％d"& m）;
printf（"输入％d转换为％s \ n，输入，dec2bin（输入，m））;
退出（0）;
}
char * dec2bin（int输入，int m）
{
static int i = 0;
char * array [m + 1];
int limit = 0;
limit = power（2，m）;
const int MASK = limit / 2; //二进制等价物是：1 + m 0位
array [m] =''\ 0'';
for（i = 0 ; i< m; i ++）
{
if（input& MASK）{
strcpy（array [i]，" 1"）;}
else {
strcpy（array [i]，" 0"）;}
input = input< < 1;
}
返回（数组）;
}
int power（int base ，int m）
{
static int i;
int p = 1;
for（i = 1; i< = m; ++ i）
{
p = p * base;
}
返回p;
}
Hi There,
I''m writing a program that takes a decimal number and returns it''s
binary equivalent.
the problem is that I don''t know how to return (from function) the
binary value as an array of cocatenated characters. For example : I
want to convert number 3 using 4 bits, ie 3 <----"0011".
After executing, I''m getting this warning:
warning: return from incompatible pointer type
line:49: warning: function returns address of local variable
Here it is my code:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int power (int base, int m);
char* dec2bin(int input, int m);
int main ()
{
int input = 0;
int m = 0;
printf("Number to convert\n");
scanf("%d", &input);
printf ("Number of bits\n");
scanf ("%d",&m);
printf("input %d is converted to %s\n",input,dec2bin(input, m));
exit (0);
}
char* dec2bin(int input, int m)
{
static int i = 0;
char* array[m+1];
int limit = 0;
limit = power(2, m);
const int MASK = limit/2; // the binary equivalent is: 1 + m 0 bits
array[m] = ''\0'';
for (i= 0; i < m ;i++)
{
if (input & MASK){
strcpy(array[i],"1");}
else{
strcpy(array[i],"0");}
input = input << 1;
}
return (array);
}
int power (int base, int m)
{
static int i;
int p=1;
for (i = 1; i <= m; ++i)
{
p =p * base;
}
return p;
}
这是最简单的修复
只能解决由提出的一个问题你的编译器：
#include< limits.h>
char * dec2bin（int input，int m）
{
static int i = 0;
静态字符数组[sizeof input * CHAR_BIT + 1];
int limit = power （2，m）;
const int MASK = limit / 2; //二进制等价物是：1 + m 0位
array [m] =''\ 0'';
for（i = 0 ; i< m; i ++）
{
if（input& MASK）{
array [i] =''1' ';
}否则{
array [i] =''0'';
}
输入=输入<< 1;
}
返回（数组）;
}
-
pete
This is the simplest fix
which only addresses the one issue raised by your compiler:
#include <limits.h>
char* dec2bin(int input, int m)
{
static int i = 0;
static char array[sizeof input * CHAR_BIT + 1];
int limit = power(2, m);
const int MASK = limit/2; // the binary equivalent is: 1 + m 0 bits
array[m] = ''\0'';
for (i= 0; i < m ;i++)
{
if (input & MASK){
array[i] = ''1'';
} else {
array[i] = ''0'';
}
input = input << 1;
}
return (array);
}
--
pete
Nezhate说：
Nezhate said:
你好，
我正在编写一个带小数的程序
Hi There,
I''m writing a program that takes a decimal number
没有这样的事情。数字是数字。十进制是位置符号
系统。数字可以用很多不同的方式表示。
No such thing. Numbers are numbers. Decimal is a positional notation
system. Numbers can be represented in lots of different ways.
并返回它的二进制等价物。
and returns it''s binary equivalent.
这是'单程：
#include< limits.h>
#include< stddef.h>
char * dec2bin（int n）
{
static char bin [sizeof n * CHAR_BIT + 1];
size_t b = sizeof n * CHAR_BIT;
size_t i = 0;
while（ b-- 0）
{
bin [i ++] =''0''+ !!（n&（1<< b））;
}
返回垃圾箱;
}
-
Richard Heathfield< http://www.cpax.org.uk>
电子邮件：-http：// www。 + rjh @
谷歌用户：< http://www.cpax.org.uk/prg/writings/googly.php>
Usenet是一个奇怪的放置" - dmr 1999年7月29日
Here''s one way:
#include <limits.h>
#include <stddef.h>
char *dec2bin(int n)
{
static char bin[sizeof n * CHAR_BIT + 1];
size_t b = sizeof n * CHAR_BIT;
size_t i = 0;
while(b-- 0)
{
bin[i++] = ''0'' + !!(n & (1 << b));
}
return bin;
}
--
Richard Heathfield <http://www.cpax.org.uk>
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999
Richard Heathfield写道：
Richard Heathfield wrote:
Nezhate说：
Nezhate said:
>嗨那里，
我正在编写一个带小数的程序
>Hi There,
I''m writing a program that takes a decimal number
没有这样的事情。数字是数字。十进制是位置符号
系统。数字可以用很多不同的方式表示。
No such thing. Numbers are numbers. Decimal is a positional notation
system. Numbers can be represented in lots of different ways.
>并返回它的二进制等价物。
>and returns it''s binary equivalent.
这里有一种方式：
#include< limits.h>
#include< stddef.h>
char * dec2bin（int n）
{
static char bin [sizeof n * CHAR_BIT + 1];
Here''s one way:
#include <limits.h>
#include <stddef.h>
char *dec2bin(int n)
{
static char bin[sizeof n * CHAR_BIT + 1];
CMIIW。如果+ 1是''\ 0''终止符，那么我们不需要那个
终结符吗？
CMIIW. If the + 1 is for the ''\0'' terminator, don''t we need that
terminator there?
size_t b = sizeof n * CHAR_BIT;
size_t i = 0;
while（b-- 0）
{
bin [i ++] =''0''+ !!（n&（1<< b））;
}
返回bin;
}
size_t b = sizeof n * CHAR_BIT;
size_t i = 0;
while(b-- 0)
{
bin[i++] = ''0'' + !!(n & (1 << b));
}
return bin;
}
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