[英]Allocating Memory to String at runtime
I am writing a program to count the occurrence of '2' followed by '1' in a sting. 我正在编写一个程序来计算字符串中出现“ 2”后跟“ 1”的次数。 I dynamically allocated string 我动态分配了字符串
Code is: 代码是:
#include <stdio.h>
#include<stdlib.h>
int penalty_shoot(char* s){
int count=0,i=0;
while(s[i]!='\0'){
if(s[i]=='2')
if(s[i+1]=='1')
count++;
i++;
}
return count;
}
int main() {
int t;
int i=0;
scanf("%d",&t); //t is for number of test cases.
while(t--){
char *str, c;
str = (char*)malloc(1*sizeof(char));
while(c = getc(stdin),c!='\n')
{
str[i] = c;
i++;
str=realloc(str,i*sizeof(char));
}
str[i] ='\0';
printf("%s\n",str);
printf("%d\n",penalty_shoot(str));
free(str);
str=NULL;
i=0;
}
return 0;
}
Input is : 输入为:
3
101201212110
10101
2120
I am facing 2 problems: 我面临2个问题:
1) I feel the dynamic allocation is not working fine.I wrote the code for dynamic allocation seeing various codes on stackoverflow . 1)我觉得动态分配不能正常工作。我在stackoverflow上看到了各种代码,为动态分配编写了代码。 (Can anyone suggest some changes.) (任何人都可以提出一些建议。)
2) The code isn't reading '2120' as the 3rd input. 2)代码没有读取“ 2120”作为第三个输入。 (why is it so ?) (为什么会这样?)
Three errors: 三个错误:
Not checking for EOF
: 不检查EOF
:
Change while(c = getc(stdin),c!='\\n')
to while(c=getc(stdin),c!='\\n'&&c!=EOF)
将while(c = getc(stdin),c!='\\n')
更改为while(c=getc(stdin),c!='\\n'&&c!=EOF)
Reallocating with the wrong number of bytes: 用错误的字节数重新分配:
Change str=realloc(str,i*sizeof(char));
更改str=realloc(str,i*sizeof(char));
to str=realloc(str,(i+1)*sizeof(char));
到str=realloc(str,(i+1)*sizeof(char));
After taking one character input we increment i
( i++
), so the next character will be stored at the ith
position. 输入一个字符后,我们将i
( i++
)递增,因此下一个字符将存储在第ith
位置。 Now, in order to store the character at ith
position, the length of the character array must be i+1
. 现在,为了将字符存储在ith
位置,字符数组的长度必须为i+1
。 So, we realloc
with i+1
. 因此,我们用i+1
realloc
。
Just for the sake of brevity, as suggested by Basile, you might as well do this: 出于简洁起见,正如Basile建议的那样,您也可以这样做:
Change
str=realloc(str,(i+1)*sizeof(char));
更改str=realloc(str,(i+1)*sizeof(char));
tostr=realloc(str,i+1);
到str=realloc(str,i+1);
Why? 为什么? Because
sizeof char
is1
byte 因为sizeof char
是1
个字节
Not consuming the \\n
after inputting t
: 输入t
后不消耗\\n
:
Change scanf("%d",&t);
更改scanf("%d",&t);
to scanf("%d ",&t);
到scanf("%d ",&t);
or scanf("%d\\n",&t);
或scanf("%d\\n",&t);
scanf("%d ",&t);
or scanf("%d\\n",&t);
或scanf("%d\\n",&t);
. 。
Either of them works. 他们中的任何一个都可以。 Why, you ask? 你为什么问? Read this explanation taken from another SO answer here : 在这里阅读从另一个SO答案中获得的解释:
An
\\n
- or any whitespace character - in the format string consumes an entire (possibly empty) sequence of whitespace characters in the input. 格式字符串中的\\n
或任何空白字符-在输入中会消耗整个(可能为空)空白字符序列。 So the scanf only returns when it encounters the next non-whitespace character, or the end of the input stream. 因此,scanf仅在遇到下一个非空白字符或输入流的结尾时才返回。
you can use scanf("%d ", &t);
您可以使用scanf("%d ", &t);
when user input to test then just before second while loop, which condition should be c != '\\n'
write c = getchar();
当用户输入要进行测试的代码时,恰好在第二次while循环之前,该条件应为c != '\\n'
编写c = getchar();
and then make sure you create a char variable, i called mine clear, that receives 0 so when you loop after initiating your string you write c = clear;
然后确保您创建一个char变量,我称为mine clear,该变量接收0,因此在启动字符串后循环时,您将代码写为c = clear;
and under it c = getchar()
again. 在其下c = getchar()
再次。 and when you use realloc make sure you make it bigger by (i+1) since char is only the size of 1 byte. 当您使用realloc时,请确保将其增大(i + 1),因为char的大小仅为1个字节。
we create the clear variable in order to clear the buffer. 我们创建clear变量以清除缓冲区。
it worked for me. 它为我工作。 make sure you insert the string all at once. 确保一次插入所有字符串。
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