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开始您需要输入多少个元素的程序,例如,如果选择了3您可以键入一个b c,但是如果输入多于3个元素a b c d该程序立即崩溃。 还没有想出如何输入超过4个元素a b c d它只会读取a b c部分。如您所愿,如果你在更多的数字比你count
进入,因为它是如何解析输入的输入数组问题
#pragma hdrstop
#pragma argsused
#include <string>
#include <tchar.h>
#include <conio.h>
#include <stdio.h>
#include <math.h>
#include <iostream.h>
#include <iomanip>
#include <sstream>
int main() {
char teikums[100]; // Masiva lielums
int c, i, count, patsk; // Patskani
char yesno; // Atkartosanas Mainigais
do {
system("cls"); // Notira Ekranu
patsk = 0; // Pieskir vertibu
cout << "Ievadi Massiva lielumu 1-100: ";
cin >> count;
if (count > 100 || count < 1) {
cout << "Massivs nedriklst but lielaks par 100 vai mazaks par 0";
}
else {
cout << "Ievadi " << count << " burtus vienu pa vienam\n";
for (i = 1; i <= count; i++) {
cin >> teikums[i];
}
cout << "\nIzmantotie Patskani:";
for (i = 0; teikums[i] != '\0'; i = i + 2) {
if (teikums[i] == 'a' || teikums[i] == 'e' ||
teikums[i] == 'o' || teikums[i] == 'o' ||
teikums[i] == 'u' || teikums[i] == 'A' ||
teikums[i] == 'E' || teikums[i] == 'I' ||
teikums[i] == 'O' || teikums[i] == 'U') {
++patsk;
cout << teikums[i];
}
}
cout << "\nPatskanu Skaits: " << patsk;
}
cout << ("\nVai velaties atkartot(Y/cits):");
// prasa lietotajam vai velas atkartot
cin >> yesno;
if (yesno == 'y' || yesno == 'Y') {
}
else {
return 0;
}
}
while (tolower(yesno) != 'n');
getch();
}
'for(i = 1; i <= count; i ++)'and'for(i = 0; teikums [i]!='\ 0'; i = i + 2)'不匹配。此外,即使程序结束与预期行为相反,它似乎也不会崩溃。 – BLUEPIXY
'teikums [i] =='o'' - >'teikums [i] =='i''''''''''''#' – BLUEPIXY
'#include''''#include ' –
BLUEPIXY