Is fadd.q integer addition?
No. It is floating-point addition with NaN, infinity, rounding, and exception-flag behavior.
Performs quad-precision FP addition: rd = rs1+rs2. R-type, fmt=Q(11).
fadd.q performs IEEE 754 addition on two quad-precision source operands, rounds according to rm, and writes the quad-precision result to f[rd].
fadd.q performs IEEE 754 addition on two quad-precision source operands, rounds according to rm, and writes the quad-precision result to f[rd].
Understand this scenario with real code like «fadd.q f0, f1, f2, rne # f0 = f1+f2».
Understand this scenario with real code like «fadd.q f0, f1, f2, rne # f0 = f1+f2».
No. It is floating-point addition with NaN, infinity, rounding, and exception-flag behavior.
No. If the result is not representable, it is rounded according to the rounding mode and may set floating-point exception flags such as NX.