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我只是從vc6 ++切換到vs2013，發現它們之間有很多不同之處。 喜歡使用scanf_s代替scanf函數，CMATH而不是文件math.h殘培？ - > _getchVS2013存在一些問題
是否有我應該知道的
順便說一句任何其他主要的區別，這是什麼意思？ 「錯誤C2668：‘戰俘’：曖昧通話超載」
這裏是我的代碼，它是一個簡單的加密程序
#include<stdio.h>
#include<string.h>
#include<conio.h>
#include<math.h>
int encode1(int d,int n);
int encode2(int d,int n);
int encode3(int d,int n);
int decode1(int d,int n);
int decode2(int d,int n);
int decode3(int d,int n);
int main(void)
{
char ans, exit;//Used to exit the program
puts("Program starts");
int d = 0, n, temp;
char s;
unsigned number;
printf("Please choose one of the followings \n");
printf("1)Encrypting\t2）Decrypting(q to quit）\n");
scanf("%d",&number);
while (number!= 1 && number != 'q'&& number != 2)
{
printf("Wrong choice, retry\n");
number = getchar();
}
if(number!='q')
{
printf("Please type in the digit of your password\n");
scanf("%d", &n);
getchar();
}
switch (number)
{
case 1: temp = encode1(d, n);
printf("%d\n", temp);
break;
case 2: temp = decode1(d, n);
printf("%d\n", temp);
break;
default:break;
}
puts("Press any key to continue");
exit = _getch();
puts("Program ends");
return 0;
}
int encode1(int d, int n)
{
int b, c[100], i, j, t, r, e[100];
puts("Please type in your password to generated the key(integers only)");
scanf("%d", &b);
for (i = 0; i<n+1; i++)
{
c[i] = b % 8;
b = b/8;
}
for (i = 0; i<n+1; i++)
{
d = d * 10 + c[i];
}
r = encode2(d,n);
return r;
}
int encode2(int b, int n)
{
int c[100], i, j, t, r, d = 0;
for (i = 0; i<n+2; i++)
{
c[i] = b % 8;
b = b/8;
}
for (i = 0; i<n+2; i++)
{
d = d * 10 + c[i];
}
r = encode3(d,n);
return r;
}
int encode3(int b, int n)
{
int c[100], i, j, t, r, d = 0;
for (i = 0; i<n+3; i++)
{
c[i] = b % 8;
b = b/8;
}
for (i = 0; i<n+3; i++)
{
d = d * 10 + c[i];
}
return d;
}
int decode1(int d, int n)
{
puts("Type in the key to retrieve your password");
scanf("%d", &d);
int a[100], t, x;
int c[100], i, j, e[100], k, g, u = 0, r;
for (i = 0, j = n+3; i<n+3, j>0; j--, i++)
{
k = pow(10, j);
e[i] = d/(k);
}
for (i = 0, j = n+2; j >= 0, i<n+3; i++, j--)
{
g = pow(10, j + 1);
k = pow(10, j);
c[i] = (d - e[i] * g)/k;
}
for (i = 0, j = n+2; i<n+3, j >= 0; i++, j--)
{
t = c[j];
a[i] = t;
}
for (i = 0, j = n+2; i<n+3, j >= 0; i++, j--)
{
r = pow(8, j);
u += a[i] * r;
}
x = decode2(u,n);
return x;
}
int decode2(int d, int n)
{
int a[100], t;
int c[100], i, j, e[100], k, g, u = 0, r, x;
for (i = 0, j = n+2; i<n+2, j>0; j--, i++)
{
k = pow(10, j);
e[i] = d/(k);
}
for (i = 0, j = n+1; j >= 0, i<n+2; i++, j--)
{
g = pow(10, j + 1);
k = pow(10, j);
c[i] = (d - e[i] * g)/k;
}
for (i = 0, j = n+1; i<n+2, j >= 0; i++, j--)
{
t = c[j];
a[i] = t;
}
for (i = 0, j = n+1; i<n+2, j >= 0; i++, j--)
{
r = pow(8, j);
u += a[i] * r;
}
x = decode3(u,n);
return x;
}
int decode3(int d, int n)
{
int a[100], t;
int c[100], i, j, e[100], k, g, u = 0, r, x;
for (i = 0, j = n+1; i<n+1, j>0; j--, i++)
{
k = pow(10, j);
e[i] = d/(k);
}
for (i = 0, j = n; j >= 0, i<n+1; i++, j--)
{
g = pow(10, j + 1);
k = pow(10, j);
c[i] = (d - e[i] * g)/k;
}
for (i = 0, j = n; i<n+1, j >= 0; i++, j--)
{
t = c[j];
a[i] = t;
}
for (i = 0, j = n; i<n+1, j >= 0; i++, j--)
{
r = pow(8, j);
u += a[i] * r;
}
return u;
}
VC6 ++？!!是的，你會遇到很多變化。如果有人有一個相當完整的名單，我會感到驚訝，但我認爲它不會受到傷害。錯誤信息告訴你'pow'有多個定義，你的調用不適合它們中的任何一個。 – 2014-10-28 00:53:00
您仍然可以使用'scanf'，關於它的警告可以忽略 – 2014-10-28 01:13:13
「模糊調用重載」表明您使用的是C++編譯器而不是C語言。 – chux 2014-10-29 05:31:41