函数无限递归，没整明白，请教

Question

这函数好象不会无限 递归吧，没整明白

function getAddTop(td_rc: { r: number, c: number, maxc: number, maxr: number }, sRow: number) {
    if (sRow < 0) return
    let td
    for (let x = 1; x < tb.rows[sRow].cells.length; x++) {
      td = tb.rows[sRow].cells[x]
      if (!td) continue
      const rc = JSON.parse(td.getAttribute('td_rc') as string)
      if (rc.maxr < td_rc.r && rc.c == td_rc.c) {
        console.log(`第${sRow}(行)次,向上增加td:${sRow}行,${x}列`, td)
        return td
      }
    }
    if (!td) { sRow = sRow - 1; getAddTop(td_rc, sRow) }
  }

Answer 1

应该是不会无限递归的,你是要返回td吗, 应该是逐层的返回

function getAddTop(td_rc: { r: number, c: number, maxc: number, maxr: number }, sRow: number) {
    if (sRow < 0) return
    let td
    for (let x = 1; x < tb.rows[sRow].cells.length; x++) {
      td = tb.rows[sRow].cells[x]
      if (!td) continue
      const rc = JSON.parse(td.getAttribute('td_rc') as string)
      if (rc.maxr < td_rc.r && rc.c == td_rc.c) {
        console.log(`第${sRow}(行)次,向上增加td:${sRow}行,${x}列`, td)
        return td
      }
    }
    return getAddTop(td_rc, sRow-1);
  }

或者改成循环,你这个不需要递归

function getAddTop(td_rc: { r: number, c: number, maxc: number, maxr: number }, sRow: number) {
    let td
    for (let y = sRow; y >=0 ; y--) {
        for (let x = 1; x < tb.rows[y].cells.length; x++) {
          td = tb.rows[y].cells[x]
          if (!td) continue
          const rc = JSON.parse(td.getAttribute('td_rc') as string)
          if (rc.maxr < td_rc.r && rc.c == td_rc.c) {
            console.log(`第${y}(行)次,向上增加td:${y}行,${x}列`, td)
            return td
          }
        }
    }
}

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