[英]Input string of unknown length
我寫了這個程序,用“ *”代替兩個空格。
如何修改代碼,以便無論字符串大小如何都能執行相同的操作? 甚至可能只使用putchar
和getchar
嗎?
#include <stdio.h>
int c;
char buffer[256];
int counter= 0;
int i;
int main()
{
while ((c = getchar()) != '\n'&&c!=EOF) {
buffer[counter] =c;
counter++;
if (counter >=255) {
break;
}
}
for(i=0; i<256; i++) {
if(buffer[i]== ' '&&buffer[i+1]==' ')
{
buffer[i]= '*';
putchar(buffer[i]);
i = i + 2;
continue;
}
putchar(buffer[i]);
}
putchar('\n');
return 0;
}
問題陳述不需要您將完整的輸入存儲在緩沖區中。 輸出什么字符的決定僅取決於輸入的最后兩個字符。 考慮以下代碼:
#include <stdio.h>
int main(void)
{
// two variables for the last two input characters
int c = EOF, last = EOF;
while ((c = getchar()) != EOF)
{
// if both are a space, store a single '*' instead
if (c == ' ' && last == ' ')
{
c = '*';
last = EOF;
}
// print the output, and shift the buffer
if (last != EOF)
putchar(last);
last = c;
}
// take care of the last character in the buffer after we see EOF
if (last != EOF)
putchar(last);
}
完全不需要malloc和朋友。 這是一個很好的示例,該問題需要您在編寫代碼之前仔細考慮,以免浪費不必要的緩沖區資源。
僅打印代碼:
#include <stdio.h>
#include <stdlib.h>
int main()
{
char prev = EOF, curr;
while ((curr =(char)getchar()) != '\n' && curr != EOF)
{
if(curr==' '&&prev==' ')
{
curr = '*';
prev = EOF;
}
if (prev != EOF)
putchar(prev);
prev = curr;
}
putchar(prev);
return 0;
}
使用realloc
實際更改字符串:
#include <stdio.h>
#include <stdlib.h>
int main() {
unsigned int len_max = 128;
char *m = malloc(len_max);
char c;
int counter= 0;
int i;
int current_size = 256;
printf("please input a string\n");
while ((c = getchar()) != '\n' && c != EOF)
{
m[counter] =(char)c;
counter++;
if(counter == current_size)
{
current_size = i+len_max;
m = realloc(m, current_size);
}
}
for(i=0; i<counter; i++)
{
if(m[i]== ' '&&m[i+1]==' ')
{
m[i]= '*';
putchar(m[i]);
i = i + 2;
continue;
}
putchar(m[i]);
}
putchar('\n');
free(m);
m = NULL;
return 0;
}
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