多人说话场景难题待解,小米开源 CocktailASR - 1 模型精准定位目标语音!
2026/9/11 23:37:16
int a = 3; int b = 5; int c = 0; c = b = a+3;// a+3=6 , b=6 ,c=b=6 从右往左依次赋值int n = 10; n = n + 5; //n += 5 15例子 ↓
int n = 10; n++; //n = n+1 → n += 1 11 //++n ,也是如此 ,n = 11int n = 10; int m = n++;// m = 10 ,n = n+1 → n = 11⚠️ 前置++ 的口诀:先自增再赋值 (前置- - 也是如此)
int n = 10; int m =++n;// n = n+1 → n = 11 → m = n = 11例子 ↓
#include<stdio.h>intmain(){inta=-5;printf("%d\n",-a);//5printf("%d\n",a);//-5//聪明的你可能已经发现,这跟数学知识是一样的💡return0;}操作符属性链接,点击一下蓝色区域,可查看
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#include<stdio.h>intmain(){charc=100;shorts=200;intr=c+s;//c 和 s 都被提升为int 后再进行加法。 r = 100 +200 =300,提升后与原值一样printf("%d\n",r);return0;}例子 ↓
#include<stdio.h>intmain(){intn=10;floatf=5.5f;floatr=n+f;//编译器自动把 n 转换为 float类型,再与 f 进行加法运算//n = 10.000000 , r = 15.500000printf("%f\n",r);return0;}形式如下:
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① int n = (int)3.14;//n = 3 ② int a = 5; int b = 2; float c = (float)a / b; //2.500000例如 ↓
a = 10; c = (b = a + 3) - (a = 5);//结果是未定义的 ① a = 5 , b = 5 + 3 = 8 , c = 3 ② b = 13 , a = 5 , c = 8例子 ↓
int i = 1; int r = (++i) + (++i) + (++i); //在不同的编译器底下,结果都不一样,不信你试试😁