小编mer*_*ito的帖子

const s: string = 'foo';

const pass1 = (origin: string) => origin.concat(s);
const pass2 = (origin: string[]) => origin.concat(s);

type S = string | string[];

const error = (origin: S) => origin.concat(s);

上面的代码。我可以调用concatastring或string[]type。那么，为什么打字稿禁止拨打concat的string | string[]类型？

错误是：

Cannot invoke an expression whose type lacks a call signature.
Type '((...strings: string[]) => string) | { (...items: ConcatArray<string>[]): string[]; (...items: (s...'
has no compatible call signatures.

因为它们有不同的返回类型？但我认为 TS 可以推断出error的类型是S. …

代码来自Kilo项目:

/*
 * Use write() and terminal escape (ESC) sequence
 * to query cursor position.
 */

#include <stdio.h>
#include <unistd.h>

#define ESC 27

int main(int argc, char const *argv[]) {
    char buf[32];
    unsigned int i = 0;
    int rows, cols;

    write(STDOUT_FILENO, "\033[6n", 4);

    while (i < sizeof(buf)-1) {
        if (read(STDIN_FILENO,buf+i,1) != 1) break;
        if (buf[i] == 'R') break;
        i++;
    }
    buf[i] = '\0';

    /* Parse it. */
    if (buf[0] != ESC || buf[1] != '[') return -1;
    if (sscanf(buf+2,"%d;%d",&rows,&cols) != …