就像 问问 你们的 VS2015 输入
#include <stdio.h>
/* count characters in input; 1st version */
main()
{
long nc;
nc = 0;
while (getchar() != EOF)
++nc;
printf("%ld\n", nc);
}
显示结果是否正常 我的是空的

应该是 C程序设计语言 里面的语法 和VS2015 不全通用 我改了
#include <stdio.h>
/* count characters in input; 1st version */
main()
{
long nc;
nc = 0;
while (getchar() != EOF)
printf("%ld\n", ++nc);
}
就能显示了

就是这 个示例
#include <stdio.h>
#define IN 1 /* inside a word */
#define OUT 0 /* outside a word */
main()
{
int c, nl, nw, nc, state;
state = OUT;
nl = nw = nc = 0;
while ((c = getchar()) != EOF) {
++nc;
if (c == '\n')
++nl;
if (c == ' ' || c == '\n' || c = '\t')
state = OUT;
else if (state == OUT) {
state = IN;
++nw;
}
}
printf("%d %d %d\n", nl, nw, nc);
}

#include <stdio.h>
#define IN 1 /* inside a word */
#define OUT 0 /* outside a word */
/* count lines, words, and characters in input */
main()
{
int c, nl, nw, nc, state;
state = OUT;
nl = nw = nc = 0;
while ((c = getchar()) != EOF) {
++nc;
if (c == '\n')
++nl;
if (c == ' ' || c == '\n' || c = '\t') 这句c == ' ' 显示 错误C2106“=”: 左操作数必须为左值
state = OUT;
else if (state == OUT) {
state = IN;
++nw;
}
}
printf("%d %d %d\n", nl, nw, nc);
}
错误(活动)表达式必须是可修改的左值