如果我5 5在终端输入,按回车键,然后再次按回车键,我想退出循环.
int readCoefficents(double complex *c){
int i = 0;
double real;
double img;
while(scanf("%f %f", &real, &img) == 2)
c[i++] = real + img * I;
c[i++] = 1 + 0*I; // most significant coefficient is assumed to be 1
return i;
}
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显然,该代码并没有为我完成工作(是的,我知道有一个缓冲区溢出等待发生).
scanf除非我输入一个字母(或一些非数字,而不是空白字符串),否则不会退出.读取空行后如何让scanf退出?
pax*_*blo 10
您遇到的具体问题是scanf格式字符串%f将跳过空格(包括换行符),直到找到要扫描的实际字符.从c99标准:
A转换规范在下面的步骤执行:
-输入的空白字符(由指定的isspace功能)被跳过,除非本说明书包括一个'[','c'或'n'说明符.
并在其他地方描述isspace():
标准的空格字符如下:空格
' ',换页'\f',换行符'\n',回车符'\r',水平制表符'\t'和垂直制表符'\v'.
最好的办法是使用fgets获取线路(这可以很容易地防止缓冲区溢出),然后sscanf在结果线上使用.
该scanf功能是您应该非常谨慎地看待的功能之一.以下代码是我经常用来处理行输入的代码:
#include <stdio.h>
#include <string.h>
#define OK 0
#define NO_INPUT 1
#define TOO_LONG 2
static int getLine (char *prmpt, char *buff, size_t sz) {
int ch, extra;
// Get line with buffer overrun protection.
if (prmpt != NULL) {
printf ("%s", prmpt);
fflush (stdout);
}
if (fgets (buff, sz, stdin) == NULL)
return NO_INPUT;
// If it was too long, there'll be no newline. In that case, we flush
// to end of line so that excess doesn't affect the next call.
if (buff[strlen(buff)-1] != '\n') {
extra = 0;
while (((ch = getchar()) != '\n') && (ch != EOF))
extra = 1;
return (extra == 1) ? TOO_LONG : OK;
}
// Otherwise remove newline and give string back to caller.
buff[strlen(buff)-1] = '\0';
return OK;
}
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// Test program for getLine().
int main (void) {
int rc;
char buff[10];
rc = getLine ("Enter string> ", buff, sizeof(buff));
if (rc == NO_INPUT) {
// Extra NL since my system doesn't output that on EOF.
printf ("\nNo input\n");
return 1;
}
if (rc == TOO_LONG) {
printf ("Input too long [%s]\n", buff);
return 1;
}
printf ("OK [%s]\n", buff);
return 0;
}
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使用各种组合进行测试:
pax> ./prog
Enter string>[CTRL-D]
No input
pax> ./prog
Enter string> a
OK [a]
pax> ./prog
Enter string> hello
OK [hello]
pax> ./prog
Enter string> hello there
Input too long [hello the]
pax> ./prog
Enter string> i am pax
OK [i am pax]
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我要做的是使用此函数安全地获取一行,然后简单地使用:
sscanf (buffer, "%f %f", &real, &img)
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获取实际值(并检查计数).
事实上,这是一个更接近你想要的完整程序:
#include <stdio.h>
#include <string.h>
#define OK 0
#define NO_INPUT 1
#define TOO_LONG 2
static int getLine (char *prmpt, char *buff, size_t sz) {
int ch, extra;
// Get line with buffer overrun protection.
if (prmpt != NULL) {
printf ("%s", prmpt);
fflush (stdout);
}
if (fgets (buff, sz, stdin) == NULL)
return NO_INPUT;
// If it was too long, there'll be no newline. In that case, we flush
// to end of line so that excess doesn't affect the next call.
if (buff[strlen(buff)-1] != '\n') {
extra = 0;
while (((ch = getchar()) != '\n') && (ch != EOF))
extra = 1;
return (extra == 1) ? TOO_LONG : OK;
}
// Otherwise remove newline and give string back to caller.
buff[strlen(buff)-1] = '\0';
return OK;
}
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int main (void) {
int i = 1, rc;
char prompt[50], buff[50];
float real, imag;
while (1) {
sprintf (prompt, "\nEnter real and imaginary for #%3d: ", i);
rc = getLine (prompt, buff, sizeof(buff));
if (rc == NO_INPUT) break;
if (*buff == '\0') break;
if (rc == TOO_LONG) {
printf ("** Input too long [%s]...\n", buff);
}
if (sscanf (buff, "%f %f", &real, &imag) == 2) {
printf ("Values were %f and %f\n", real, imag);
i++;
} else {
printf ("** Invalid input [%s]\n", buff);
}
}
return 0;
}
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以及测试运行:
pax> ./testprog
Enter real and imaginary for # 1: hello
** Invalid input [hello]
Enter real and imaginary for # 1: hello there
** Invalid input [hello there]
Enter real and imaginary for # 1: 1
** Invalid input [1]
Enter real and imaginary for # 1: 1.23 4.56
Values were 1.230000 and 4.560000
Enter real and imaginary for # 2:
pax> _
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使用fgets读取控制台输入:
int res = 2;
while (res == 2) {
char buf[100];
fgets(buf, sizeof(buf), stdin);
res = sscanf(buf, "%f %f", &real, &img);
if (res == 2)
c[i++] = real + img * I;
}
c[i++] = 1 + 0*I; // most significant coefficient is assumed to be 1
return i;
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